---
title: "Integration"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 9
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/9-integration
---

# Chapter 9 — Integration

The Year 1 volume built the integral on a segment. This chapter extends it to arbitrary intervals ([improper integrals](#def-b2-integration-improper), with the full comparison toolkit), then studies integrals *depending on a parameter* — [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and differentiation under the integral sign — powered by the dominated convergence theorem, the one result of this chapter taken on trust. The $\Gamma$ function serves as the running example, and as the gateway to half of the special functions of mathematics.

## 9.1 Integrals on an arbitrary interval

**Definition 9.1.**

Let $f$ be piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intco{a}{b}$ ($b \in \R$ or $+\infty$). The integral *converges* when $\lim_{x \to b^-} \int_a^x f$ exists; one then writes $\int_a^b f$ for the limit. (Similarly on $\intoc{a}{b}$, and on $\intoo{a}{b}$ by splitting at an interior point — the choice does not matter, by Chasles.) The integral *converges [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def)* when $\int_a^b \abs f$ converges; absolute convergence implies convergence, by the Cauchy criterion:

$$
\Bigl| \int_x^{y} f \Bigr| \leq \int_x^{y} \abs f
$$

and [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) of $\R$ (the primitive has the Cauchy property). In detail: let $F(x) = \int_a^x f$ and $G(x) = \int_a^x \abs
f$. If $\int^b\abs f$ converges, $G$ has a limit at $b^-$, so for every $\varepsilon > 0$ there is $c < b$ with $G(y) - G(x)
\leq \varepsilon$ whenever $c \leq x \leq y < b$; the display transfers this Cauchy property to $F$. For any sequence $x_n
\to b^-$ the values $F(x_n)$ then form a Cauchy sequence of reals, convergent by [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete), and interlacing two such sequences shows the limit is the same for all of them: $F$ has a limit at $b^-$.

**Theorem 9.2 (Positive comparison toolkit).**

For $f, g \geq 0$ piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intco{a}{b}$:

1. $\int_a^b f$ [converges](#def-b2-integration-improper) iff the primitive $x \mapsto \int_a^x  f$ is bounded;
2. $f \leq g$ : convergence of $\int g$ forces that of $\int f$ ; divergence transfers the other way;
3. $f \sim g$ at $b$ : the two integrals have the same nature;
4. the reference scales: at $+\infty$ , $\int^{\infty}  \frac{\dd t}{t^\alpha}$ [converges](#def-b2-integration-improper) iff $\alpha > 1$ , and $\int^\infty \frac{\dd t}{t(\ln t)^\beta}$ iff $\beta > 1$ ; at a finite endpoint $b$ , $\int^b \frac{\dd t}{(b -  t)^\alpha}$ [converges](#def-b2-integration-improper) iff $\alpha < 1$ .

**Proof.** (1) The primitive $F(x) = \int_a^x f$ is nondecreasing ($f \geq
0$). If it is bounded, $\ell = \sup_{x < b}F$ is finite and $F(x) \to \ell$: given $\varepsilon > 0$, some $F(x_0) > \ell -
\varepsilon$, and monotonicity traps $F(x) \in
\intoc{\ell - \varepsilon}{\ell}$ for $x_0 \leq x < b$. If it is unbounded, $F \to +\infty$: divergence.

(2) From $f \leq g$: $\int_a^x f \leq \int_a^x g$ for all $x$; if $\int^b g$ [converges](#def-b2-integration-improper), the right side is bounded, hence so is the left, and (1) concludes. Contraposition transfers divergence the other way.

(3) $f \sim g$ at $b$ provides $c < b$ with

$$
\tfrac12\,g(t) \;\leq\; f(t) \;\leq\; 2\,g(t)
\qquad (c \leq t < b) :
$$

by (2) applied both ways on $\intco{c}{b}$, the two integrals have the same nature; the initial piece $\intcc{a}{c}$ is a proper integral and changes nothing.

(4) Explicit primitives: for $\alpha \neq 1$ and $\beta \neq 1$,

$$
\int_c^x \frac{\dd t}{t^\alpha}
= \frac{x^{1-\alpha} - c^{1-\alpha}}{1 - \alpha},
\qquad
\int_c^x \frac{\dd t}{t(\ln t)^\beta}
= \frac{(\ln x)^{1-\beta} - (\ln c)^{1-\beta}}{1 - \beta},
$$

with logarithms in the excluded cases: bounded as $x \to
+\infty$ exactly when $\alpha > 1$, resp. $\beta > 1$. At a finite endpoint, the substitution $u = b - t$ reduces to the scale $\int_0 u^{-\alpha}\,\dd u$, bounded iff $\alpha < 1$. Apply (1) each time. ∎

**Example 9.3 (Two warm-ups, worked to the end).**

*(a)* $\displaystyle\int_0^1 \ln t\,\dd t$: the integrand blows up at $0^+$, but $\abs{\ln t} = o\bigl(t^{-1/2}\bigr)$ there (logarithms lose to powers), and $\int_0 t^{-1/2}$ [converges](#def-b2-integration-improper): absolute convergence. The value, by parts on $\intcc{\varepsilon}{1}$:

$$
\int_\varepsilon^1 \ln t\,\dd t
= \bigl[t\ln t - t\bigr]_\varepsilon^1
= -1 - \varepsilon\ln\varepsilon + \varepsilon
\xrightarrow[\varepsilon\to0^+]{} -1 .
$$

*(b)* $\displaystyle\int_0^\infty \frac{\ln t}{1 +
t^2}\,\dd t$: trouble at both ends, so split at $1$. Near $0$: $\abs{\ln t}$ integrable as in (a); near $\infty$: $\frac{\ln
t}{1+t^2} = o(t^{-3/2})$: [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent. The substitution $t = \frac1u$ maps $\intoo{0}{1}$ onto $\intoo{1}{\infty}$ and

$$
\int_0^1 \frac{\ln t}{1+t^2}\,\dd t
= \int_1^{\infty} \frac{-\ln u}{1 + u^{-2}}\cdot
\frac{\dd u}{u^2}
= -\int_1^\infty \frac{\ln u}{1+u^2}\,\dd u :
$$

the two halves cancel, and the integral is $0$. Closing insight: symmetry under $t \mapsto \frac1t$ is worth a page of computation — the same trick already powered [Exercise 9.3](#exo-b2-integration-3).

**Example 9.4 (One value, three integrals).**

Study $I = \displaystyle\int_0^{\infty} \frac{1 - \cos
t}{t^2}\,\dd t$. At $0$: $1 - \cos t \sim \frac{t^2}2$, so the integrand extends [continuously](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) by the value $\frac12$ — no singularity at all. At $\infty$: $0 \leq \frac{1 - \cos t}{t^2}
\leq \frac{2}{t^2}$: absolute convergence ([Theorem 9.2](#thm-b2-integration-comparison)). Value: integrate by parts on $\intcc{\varepsilon}{M}$ with $u = 1 - \cos t$, $v' =
t^{-2}$:

$$
\int_\varepsilon^M \frac{1 - \cos t}{t^2}\,\dd t
= \Bigl[-\frac{1 - \cos t}{t}\Bigr]_\varepsilon^M
+ \int_\varepsilon^M \frac{\sin t}{t}\,\dd t .
$$

The bracket vanishes at both ends ($\frac{1 -
\cos\varepsilon}{\varepsilon} \sim \frac\varepsilon2$; bounded numerator at $M$), and the integral tends to the Dirichlet value $\frac\pi2$ ([Exercise 9.10](#exo-b2-integration-10)): $I = \frac\pi2$. Closing insight: with $1 - \cos t = 2\sin^2\frac t2$ and $u =
\frac t2$,

$$
I = \int_0^\infty \frac{2\sin^2 u}{(2u)^2}\;2\,\dd u
= \int_0^\infty \Bigl(\frac{\sin u}{u}\Bigr)^{\!2}\dd u :
$$

the three classics $\int_0^\infty\frac{\sin t}{t}\dd t$, $\int_0^\infty\bigl(\frac{\sin t}{t}\bigr)^2\dd t$ ([Exercise 9.11](#exo-b2-integration-11)) and $I$ all share the value $\frac\pi2$, passed around by parts and substitution — and only the first is semi-convergent: integration by parts traded away the absolute convergence for a simpler integrand.

**Example 9.5 (A semi-convergent integral).**

$\displaystyle\int_1^{\infty} \frac{\sin t}{t}\,\dd t$ [converges](#def-b2-integration-improper): integrate by parts,

$$
\int_1^x \frac{\sin t}{t}\dd t
= \Bigl[\frac{-\cos t}{t}\Bigr]_1^x - \int_1^x \frac{\cos
t}{t^2}\dd t ,
$$

where the bracket has a limit and the last integral [converges](#def-b2-integration-improper) [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) ($\abs{\cos t}/t^2 \leq t^{-2}$). But not [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def): from $\abs{\sin t} \geq \sin^2 t$,

$$
\int_1^x \frac{\abs{\sin t}}{t}\,\dd t
\;\geq\; \int_1^x \frac{\sin^2t}{t}\,\dd t
= \underbrace{\int_1^x \frac{\dd t}{2t}}_{=\ \frac12\ln x
\ \to\ \infty}
\;-\; \underbrace{\int_1^x \frac{\cos 2t}{2t}\,\dd
t}_{\text{convergent}} ,
$$

where the last integral [converges](#def-b2-integration-improper) by the *same* integration by parts as above (with $\sin 2t$ in the bracket): a divergent piece minus a convergent one diverges. So $\int_1^\infty\frac{\sin t}{t}\dd t$ [converges](#def-b2-integration-improper) without converging [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) — the integral analogue of the alternating series, with integration by parts playing the role of the alternating test.

## 9.2 The convergence theorem

**Theorem 9.6 (Dominated convergence).**

Let $(f_n)$ be piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on an interval $I$, converging pointwise to a piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$, and suppose there is a fixed integrable $\varphi \geq 0$ ($\int_I \varphi < \infty$) with

$$
\abs{f_n(t)} \leq \varphi(t) \qquad (t \in I,\ n \in \N).
$$

Then all $\int_I f_n$ and $\int_I f$ converge [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def), and

$$
\int_I f_n \xrightarrow[n \to \infty]{} \int_I f .
$$

**Proof.** *Admitted at this level.* ∎

**Remark 9.7.**

The honest proof belongs to the Lebesgue integration theory of Year 3; the statement, however, is used constantly from now on. The *domination* hypothesis is the whole point: pointwise convergence alone does not suffice ($f_n = n\,\mathbf{1}_{\intoo{0}{1/n}}$, sliding bumps: $\int f_n = 1 \not\to 0 = \int f$). The theorem also holds for a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) parameter ($f_\lambda$, $\lambda \to
\lambda_0$), by the sequential characterization of limits.

**Example 9.8 (A Gaussian limit, by domination).**

Compute $\displaystyle\lim_{n\to\infty} I_n$ where $I_n =
\int_0^\infty \Bigl(1 + \frac{t^2}{n}\Bigr)^{\!-n}\dd t$. Pointwise, $(1 + t^2/n)^n \to \eu^{t^2}$ (compound-interest limit), so the integrands tend to $\eu^{-t^2}$. Domination: the sequence $n \mapsto (1 + u/n)^n$ is nondecreasing for $u \geq 0$ (AM–GM on the $n + 1$ factors $1, 1 + \frac un, \dots, 1 +
\frac un$ gives $(1 + \frac u{n+1})^{n+1} \geq (1 + \frac
un)^n$), so for $n \geq 2$:

$$
\Bigl(1 + \frac{t^2}{n}\Bigr)^{\!-n}
\leq \Bigl(1 + \frac{t^2}{2}\Bigr)^{\!-2},
$$

an integrable dominator ($\sim 4t^{-4}$ at infinity). Dominated convergence:

$$
I_n \xrightarrow[n\to\infty]{} \int_0^\infty \eu^{-t^2}\dd t
= \frac{\sqrt\pi}{2}
$$

(the Gaussian integral of [Exercise 9.8](#exo-b2-integration-8)). Closing check: the substitution $t = \sqrt n\tan\theta$ computes $I_n$ exactly, $I_n = \sqrt n\int_0^{\pi/2}\cos^{2n-2}\theta\,\dd\theta
= \sqrt n\,W_{2n-2}$, and the Wallis asymptotics $W_m \sim
\sqrt{\pi/(2m)}$ ([Lemma 6.11](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#lem-b2-comparison-wallis)) give $\sqrt
n\,W_{2n-2} \to \frac{\sqrt\pi}2$ again: the two pillars of this chapter and the last agree.

**Example 9.9 (Dominated convergence, continuous parameter).**

Compute

$$
\lim_{x\to+\infty}\int_0^\infty
\frac{\arctan(xt)}{1+t^2}\,\dd t .
$$

For each $t > 0$, $\arctan(xt) \to \frac\pi2$ as $x \to \infty$; and the domination

$$
\Bigl|\frac{\arctan(xt)}{1+t^2}\Bigr|
\leq \frac{\pi/2}{1+t^2},
\qquad\text{integrable, independent of } x,
$$

holds for all $x$. By the continuous-parameter form of [Theorem 9.6](#thm-b2-integration-dominated) (sequential characterization: test along every $x_n \to \infty$),

$$
\int_0^\infty\frac{\arctan(xt)}{1+t^2}\,\dd t
\xrightarrow[x\to+\infty]{}
\frac\pi2\int_0^\infty\frac{\dd t}{1+t^2}
= \frac{\pi^2}{4} .
$$

Closing insight: the single point $t = 0$, where the pointwise limit is $0$ rather than $\frac\pi2$, changes nothing — the limit function only enters through its integral, one of the quiet mercies of the theorem.

## 9.3 Integrals with a parameter

**Theorem 9.10 (Continuity under the integral sign).**

Let $f \colon A \times I \to \R$ ($A$ a [metric space](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-def), $I$ an interval) with: $t \mapsto f(x, t)$ piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) for each $x$; $x \mapsto f(x, t)$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) for each $t$; and a *domination* $\abs{f(x,t)} \leq \varphi(t)$ ($\varphi$ integrable on $I$, independent of $x$). Then

$$
F(x) = \int_I f(x, t)\,\dd t
$$

is defined and [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $A$.

**Proof.** Definedness: domination gives absolute convergence. [Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $x_0$: for any sequence $x_n \to x_0$, the functions $g_n(t) =
f(x_n, t)$ converge pointwise to $f(x_0, t)$ ([continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $x$) under the fixed domination $\varphi$: dominated convergence gives $F(x_n) \to F(x_0)$; conclude by the sequential characterization of [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Definition 4.5](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)). ∎

**Theorem 9.11 (Differentiation under the integral sign).**

Let $f \colon J \times I \to \R$ ($J$ an interval of parameters) with: $t \mapsto f(x,t)$ integrable on $I$ for each $x$; $x \mapsto
f(x,t)$ of class $C^1$ for each $t$, the partial derivative $\frac{\partial f}{\partial x}$ being piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $t$ and dominated: $\bigl|\frac{\partial f}{\partial x}(x,t)\bigr| \leq
\psi(t)$ with $\psi$ integrable. Then $F(x) = \int_I f(x,t)\dd t$ is $C^1$ on $J$ and

$$
F'(x) = \int_I \frac{\partial f}{\partial x}(x, t)\,\dd t .
$$

**Proof.** Fix $x$ and $h_n \to 0$. The difference quotients

$$
\frac{F(x + h_n) - F(x)}{h_n}
= \int_I \frac{f(x + h_n, t) - f(x, t)}{h_n}\,\dd t
$$

have integrands converging pointwise to $\frac{\partial f}{\partial
x}(x, t)$, and dominated by $\psi(t)$: by the mean value inequality applied in $x$ at fixed $t$,

$$
\Bigl|\frac{f(x + h_n, t) - f(x,t)}{h_n}\Bigr|
\leq \sup_{\xi} \Bigl|\frac{\partial f}{\partial x}(\xi, t)\Bigr|
\leq \psi(t) .
$$

Dominated convergence gives the limit $\int_I \frac{\partial
f}{\partial x}(x,t)\dd t$ of the quotients: $F$ is differentiable with the announced derivative, which is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) by [Theorem 9.10](#thm-b2-integration-continuity) applied to $\frac{\partial
f}{\partial x}$. ∎

**Example 9.12 (A parameter integral checked against a formula).**

Let $F(x) = \displaystyle\int_0^\infty \frac{\dd t}{t^2 + x}$ for $x > 0$. On every $\intcc{a}{b} \subset \intoo{0}{\infty}$, the integrand is dominated by $\frac{1}{t^2 + a}$, integrable and independent of $x$: $F$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 9.10](#thm-b2-integration-continuity)). Here the theorem can be checked against an explicit value:

$$
F(x) = \Bigl[\frac{1}{\sqrt x}\arctan\frac{t}{\sqrt
x}\Bigr]_0^\infty = \frac{\pi}{2\sqrt x} ,
$$

visibly [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). Now differentiate under the integral: the $x$-derivative $-\frac{1}{(t^2+x)^2}$ is dominated on $\intcc ab$ by $\frac{1}{(t^2+a)^2}$, integrable: [Theorem 9.11](#thm-b2-integration-leibnizrule) gives

$$
F'(x) = -\int_0^\infty \frac{\dd t}{(t^2 + x)^2}
\qquad\text{while}\qquad
F'(x) = -\frac{\pi}{4}\,x^{-3/2} ,
$$

so we have *computed* a new integral for free: $\int_0^\infty\frac{\dd t}{(t^2+x)^2} = \frac{\pi}{4x^{3/2}}$. Closing insight: differentiating a known [parameter integral](#thm-b2-integration-continuity) is a factory of new formulas — iterating gives $\int_0^\infty\frac{\dd t}{(t^2+1)^n}$ for every $n$, with no trigonometric substitutions.

**Method 9.13 (Studying an improper integral).**

Given $\int_a^b f$:

1. Locate the trouble: list the endpoints (or interior points) where $f$ is unbounded or the interval is infinite, and split so that each piece has exactly one troublesome end.
2. If $f$ has constant sign near that end, find an equivalent and compare with the reference scales of [Theorem 9.2](#thm-b2-integration-comparison) .
3. If $f$ oscillates, test $\abs f$ first (absolute convergence). If $\int\abs f$ diverges, try integration by parts to trade the oscillation for decay, as in [Example 9.5](#ex-b2-integration-sint) ; minorations like $\abs{\sin t} \geq \sin^2t$ detect genuine semi-convergence.
4. For a value, not just the nature: parts, substitution, or a parameter (differentiate a simpler integral, as in [Example 9.12](#ex-b2-integration-arctanparam) and [Example 9.21](#ex-b2-integration-laplace) ).
5. Sanity checks on any computed value: sign and rough size against a crude bound ( $\int_0^\infty  \eu^{-t^2}\dd t \in \intoo{0}{1 + \int_1^\infty  \eu^{-t}}$ , so $\frac{\sqrt\pi}{2} \approx 0.886$ is plausible); and dimensional consistency under scaling ( $t \mapsto \lambda t$ must rescale both sides the same way — the fastest detector of a lost factor).

**Remark 9.14 (Common pitfalls).**

Three recurring errors. *(i) Parameter-dependent dominators:* the domination $\abs{f(x,t)} \leq \varphi(t)$ must be uniform in $x$ on the set considered; it usually holds on segments $\intcc ab$ but not globally — for $\int_0^\infty\eu^{-xt}\dd t$ there is no integrable dominator valid for all $x > 0$, yet dominating on $x \geq a > 0$ is enough to work on the whole [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) half-line, since [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and derivatives are local notions. *(ii) Comparing signed integrands:* the comparison toolkit is for *nonnegative* functions; from $\abs f \leq g$ with $\int g$ divergent one may conclude nothing — $\int_1^\infty\frac{\sin t}t\,\dd t$ [converges](#def-b2-integration-improper) although every comparison with $\frac1t$ fails. *(iii) Forgetting half the trouble:* on $\intoo{0}{\infty}$ always study both ends separately; $\int_0^\infty\frac{\dd t}{t}$ diverges at *both*, and a convergent-looking split can silently cancel two infinities. The safe reflex is the checklist of [Method 9.13](#met-b2-integration-study).

**Example 9.15 (A Bertrand boundary case, to the digit).**

The scale $\int^\infty\frac{\dd t}{t(\ln t)^\beta}$ of [Theorem 9.2](#thm-b2-integration-comparison) sits exactly on the edge of the power scales; its boundary cases deserve one full computation. For $\beta = 2$:

$$
\int_\eu^{\infty}\frac{\dd t}{t(\ln t)^2}
= \Bigl[-\frac{1}{\ln t}\Bigr]_\eu^{\infty}
= 0 - (-1) = 1 ,
$$

a convergent integral with a pleasantly exact value; while for $\beta = 1$,

$$
\int_\eu^{x}\frac{\dd t}{t\ln t}
= \bigl[\ln\ln t\bigr]_\eu^{x} = \ln\ln x
\longrightarrow \infty ,
$$

divergent — but so slowly that reaching $\ln\ln x = 10$ requires $x = \eu^{\eu^{10}} \approx 10^{9566}$. Closing insight: between “every power $t^{-1-\varepsilon}$ [converges](#def-b2-integration-improper)” and “$t^{-1}$ diverges” lives an infinite ladder of logarithmic scales, each refining the last; the substitution $u = \ln t$ collapses each rung onto the previous one, which is why the Bertrand criteria echo the Riemann ones one level up.

**Remark 9.16 (Perspectives within this volume).**

This chapter’s tools are about to be everywhere. Dominated convergence is the engine behind the approximate identities of the next chapter (sliding kernels, Bernstein and Fejér alike); [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and differentiation under the integral sign produce the Fourier coefficients’ calculus in the Fourier chapter, where every $c_n(f)$ is a [parameter integral](#thm-b2-integration-continuity) in disguise. The $\Gamma$ function returns twice: in the chapter on multiple integrals, where a double integral finally proves Euler’s Beta–Gamma formula in full, and in the probability chapters, where $\Gamma$-type integrals normalize the standard densities and compute their moments. And the semi-convergent $\int\frac{\sin t}{t}$ resurfaces as the Gibbs constant of the Fourier chapter — the same integral, measuring the overshoot of partial sums at a jump.

**Definition 9.17 (The Γ\GammaΓ function).**

For $x > 0$:

$$
\Gamma(x) = \int_0^{\infty} t^{x-1}\,\eu^{-t}\,\dd t ,
$$

convergent at both ends ($t^{x-1}$ integrable at $0^+$ for $x > 0$; exponential decay at $\infty$).

**Theorem 9.18.**

$\Gamma$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intoo{0}{+\infty}$, satisfies the functional equation

$$
\Gamma(x + 1) = x\,\Gamma(x),
\qquad \Gamma(1) = 1,
\qquad\text{hence}\qquad \Gamma(n + 1) = n! ,
$$

and is of class $C^1$ (indeed $C^\infty$) with $\Gamma'(x) =
\int_0^\infty t^{x-1}\eu^{-t}\ln t\,\dd t$.

**Proof.** Functional equation: integrate by parts on $\intcc{\varepsilon}{M}$ and let the ends go: $\int t^{x}\eu^{-t} = [-t^x\eu^{-t}] + x\int
t^{x-1}\eu^{-t}$, boundary terms vanishing — indeed $\varepsilon^x\eu^{-\varepsilon} \to 0$ as $\varepsilon \to
0^+$ because $x > 0$, and $M^x\eu^{-M} \to 0$ as $M \to \infty$ because the exponential beats every power; both truncated integrals converge to their improper values by the convergence established in [Definition 9.17](#def-b2-integration-gamma). $\Gamma(1) = \int
\eu^{-t} = 1$; induction gives the factorial.

[Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intcc{a}{b} \subset \intoo{0}{\infty}$: dominate $t^{x-1}\eu^{-t}$ by $\varphi(t) = (t^{a-1} + t^{b-1})\eu^{-t}$, integrable and independent of $x \in \intcc{a}{b}$: [Theorem 9.10](#thm-b2-integration-continuity) applies on every such segment, hence on the whole half-line. Differentiability: the $x$-derivative $t^{x-1}\eu^{-t}\ln t$ is dominated on $\intcc{a}{b}$ by $(t^{a-1}
+ t^{b-1})\eu^{-t}\,\abs{\ln t}$, still integrable: [Theorem 9.11](#thm-b2-integration-leibnizrule); iterating gives all derivatives (each adds a power of $\ln t$, harmless). ∎

**Example 9.19 (Half-integer factorials).**

The functional equation and $\Gamma\bigl(\frac12\bigr) =
\sqrt\pi$ (a substitution away from [Exercise 9.8](#exo-b2-integration-8): set $t = u^2$ in the defining integral) generate all half-integer values:

$$
\Gamma\Bigl(\frac32\Bigr) = \frac12\,\Gamma\Bigl(\frac12\Bigr)
= \frac{\sqrt\pi}{2},
\qquad
\Gamma\Bigl(\frac52\Bigr) = \frac32\cdot\frac{\sqrt\pi}{2}
= \frac{3\sqrt\pi}{4},
\qquad
\Gamma\Bigl(\frac72\Bigr) = \frac{15\sqrt\pi}{8} .
$$

Since $\Gamma(n+1) = n!$, it is fair to say “$\frac12! =
\frac{\sqrt\pi}{2} \approx 0.886$”: the factorial has been interpolated, and the interpolating curve dips below $1$ between $0! = 1$ and $1! = 1$ (its minimum $\approx 0.8856$ at $x \approx 1.4616$ matches the convexity picture of the weekend problem’s Part I). Closing insight: nothing in the integral $\int_0^\infty t^{x-1}\eu^{-t}\dd t$ privileges integers — the factorial’s discreteness was an accident of counting, and $\sqrt\pi$ is what lives between $1$ and $1$.

**Remark 9.20 (Where Γ\GammaΓ goes from here).**

The weekend problem of this chapter builds the whole Euler calculus around $\Gamma$: the [Beta function](#pb-b2-integration-1), its integration-by-parts recursions, the Wallis integrals as Beta values, and Gauss’s limit formula. The chapter on multiple integrals proves Euler’s Beta–Gamma formula for all arguments by a double integral; the probability chapters meet $\Gamma$ again in the normalization of the most common densities and in the moments of waiting times. The Year 3 volume rebuilds $\Gamma$ on Lebesgue foundations, proves the Bohr–Mollerup uniqueness theorem, and extends Stirling’s formula from integers to the real half-line by dominated convergence.

**Example 9.21 (A classical computation by differentiation).**

For $x \in \R$, let $F(x) = \int_0^{\infty} \eu^{-t^2}\cos(xt)\,\dd
t$ ([absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent, dominated by $\eu^{-t^2}$). By [Theorem 9.11](#thm-b2-integration-leibnizrule) (domination of the $x$-derivative by $t\,\eu^{-t^2}$, integrable):

$$
F'(x) = -\int_0^\infty t\,\eu^{-t^2}\sin(xt)\,\dd t
= \Bigl[\frac{\eu^{-t^2}}{2}\sin(xt)\Bigr]_0^\infty
- \frac x2\int_0^\infty \eu^{-t^2}\cos(xt)\,\dd t
= -\frac x2\,F(x),
$$

(parts with $u' = t\eu^{-t^2}$). The differential equation $F' =
-\frac x2 F$ integrates to $F(x) = F(0)\,\eu^{-x^2/4}$: the Gaussian-type integral reproduces itself. The constant $F(0) =
\int_0^\infty \eu^{-t^2}\dd t = \frac{\sqrt\pi}{2}$ is computed in [Exercise 9.8](#exo-b2-integration-8) — and again, by double integration, in [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint).

## 9.4 Exercises

**Exercise 9.1 ★.**

Nature of: $\displaystyle\int_0^1 \frac{\dd t}{\sqrt{t(1-t)}}$; $\;\displaystyle\int_1^\infty \frac{\ln t}{t^2}\dd t$; $\;\displaystyle\int_0^\infty \frac{\dd t}{1 + t^2\sin^2 t}$ *(compare with the divergent harmonic-type behavior near $t =
n\pi$)*.

**Solution of Exercise 9.1.**

$\int_0^1 \frac{\dd t}{\sqrt{t(1-t)}}$: near $0$, $\sim t^{-1/2}$ ($\alpha = \frac12 < 1$: [converges](#def-b2-integration-improper)); near $1$, $\sim
(1-t)^{-1/2}$: [converges](#def-b2-integration-improper). Convergent (its value is $\pi$, by the substitution $t = \sin^2\theta$).

$\int_1^\infty \frac{\ln t}{t^2}$: $\frac{\ln t}{t^2} = o(t^{-3/2})$: convergent (value $1$ by parts).

$\int_0^\infty \frac{\dd t}{1 + t^2\sin^2 t}$: divergent. Near $t =
n\pi$, write $t = n\pi + u$: $\sin^2 t = \sin^2 u \leq u^2$, so on $\abs u \leq \frac{1}{n}$, $1 + t^2\sin^2 t \leq 1 + (n\pi +
1)^2u^2 \leq C n^2 u^2 + 1$; hence

$$
\int_{n\pi - 1/n}^{n\pi + 1/n} \frac{\dd t}{1 + t^2\sin^2 t}
\geq \int_{-1/n}^{1/n} \frac{\dd u}{1 + Cn^2u^2}
= \frac{2\arctan\sqrt C}{\sqrt C}\cdot\frac{1}{n} ,
$$

a term of a divergent harmonic-type series: summing over $n$, the primitive is unbounded.

**Exercise 9.2 ★.**

Compute $\displaystyle\int_0^\infty t^n \eu^{-\lambda t}\,\dd t$ ($\lambda > 0$) via $\Gamma$, and $\displaystyle\int_0^1 (\ln t)^n \dd t$ via the substitution $t =
\eu^{-u}$.

**Solution of Exercise 9.2.**

Substitute $u = \lambda t$:

$$
\int_0^\infty t^n \eu^{-\lambda t}\dd t
= \frac{1}{\lambda^{n+1}}\int_0^\infty u^n\eu^{-u}\dd u
= \frac{\Gamma(n+1)}{\lambda^{n+1}} = \frac{n!}{\lambda^{n+1}} .
$$

With $t = \eu^{-u}$ ($\dd t = -\eu^{-u}\dd u$):

$$
\int_0^1 (\ln t)^n \dd t = \int_0^{\infty} (-u)^n \eu^{-u}\,\dd u
= (-1)^n\, n! .
$$

**Exercise 9.3 ★.**

Prove that $\displaystyle\int_0^{\infty} \frac{\dd t}{(1 +
t^2)(1 + t^x)}$ is well defined for every $x \in \R$ and independent of $x$. *(Substitute $t \mapsto \frac1t$ and average the two expressions.)* What is its value?

**Solution of Exercise 9.3.**

Convergence: the integrand is $\leq \frac{1}{1+t^2}$ near $\infty$ and bounded near $0$ (both factors bounded below away from $0$): [absolutely](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-def) convergent, for every $x$. Substituting $t = \frac1u$ ($\dd t = -\frac{\dd u}{u^2}$):

$$
I(x) = \int_0^\infty \frac{1}{\bigl(1 +
\frac1{u^2}\bigr)\bigl(1 + u^{-x}\bigr)}\cdot\frac{\dd u}{u^2}
= \int_0^\infty \frac{u^x}{(1 + u^2)(1 + u^x)}\,\dd u .
$$

Adding the two expressions of $I(x)$:

$$
2I(x) = \int_0^\infty \frac{1 + t^x}{(1+t^2)(1+t^x)}\dd t
= \int_0^\infty \frac{\dd t}{1 + t^2} = \frac{\pi}{2} :
$$

$I(x) = \frac\pi4$, independent of $x$.

**Exercise 9.4 ★★.**

(Bertrand integrals at a finite endpoint) For which $(\alpha,
\beta)$ does $\displaystyle\int_0^{1/2}
\frac{\dd t}{t^\alpha\,\abs{\ln t}^\beta}$ converge?

**Solution of Exercise 9.4.**

Near $0^+$, with $u = \abs{\ln t} \to \infty$. If $\alpha < 1$: convergence regardless of $\beta$ (compare with $t^{-\alpha'}$ for $\alpha < \alpha' < 1$: the log factor is beaten). If $\alpha > 1$: divergence regardless of $\beta$ (compare with $t^{-\alpha''}$, $1 < \alpha'' < \alpha$). If $\alpha = 1$: substitute $t =
\eu^{-u}$:

$$
\int_0^{1/2} \frac{\dd t}{t\,\abs{\ln t}^\beta}
= \int_{\ln 2}^{\infty} \frac{\dd u}{u^\beta},
$$

convergent iff $\beta > 1$. Summary: convergence iff $\alpha < 1$, or ($\alpha = 1$ and $\beta > 1$) — the mirror of the Bertrand series.

**Exercise 9.5 ★★.**

Let $F(x) = \displaystyle\int_0^{\infty} \frac{\eu^{-xt}}{1 +
t^2}\,\dd t$ for $x \geq 0$. Prove that $F$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intco{0}{\infty}$, $C^2$ on $\intoo{0}{\infty}$, satisfies $F'' +
F = \frac1x$ there, and that $F(x) \to 0$ as $x \to +\infty$.

**Solution of Exercise 9.5.**

[Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intco{0}{\infty}$: domination $\bigl|\frac{\eu^{-xt}}{1+t^2}\bigr| \leq \frac{1}{1+t^2}$, integrable, uniform in $x \geq 0$: [Theorem 9.10](#thm-b2-integration-continuity).

$C^2$ on $\intoo{0}{\infty}$: on $x \geq a > 0$, the first two $x$-derivatives $\frac{-t\,\eu^{-xt}}{1+t^2}$ and $\frac{t^2\eu^{-xt}}{1+t^2}$ are dominated by $t\,\eu^{-at}$ and $\eu^{-at}$: two applications of [Theorem 9.11](#thm-b2-integration-leibnizrule). Then

$$
F''(x) + F(x) = \int_0^\infty \frac{t^2 + 1}{1 + t^2}\,\eu^{-xt}\dd
t = \int_0^\infty \eu^{-xt}\dd t = \frac1x .
$$

Limit: $0 \leq F(x) \leq \int_0^\infty \eu^{-xt}\dd t = \frac1x \to
0$.

**Exercise 9.6 ★★.**

(Frullani) Let $f$ be [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $\intco{0}{+\infty}$ with a finite limit $f(\infty)$ at $+\infty$. Prove that for $a, b > 0$:

$$
\int_0^{\infty} \frac{f(at) - f(bt)}{t}\,\dd t
= \bigl(f(0) - f(\infty)\bigr)\,\ln\frac ba .
$$

*(On $\intcc{\varepsilon}{M}$, substitute in each piece and regroup into $\int_{a\varepsilon}^{b\varepsilon} -
\int_{aM}^{bM}$ of $\frac{f(u)}u\,\dd u$; squeeze using the [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$ and the limit at $\infty$.)* Compute $\int_0^\infty \frac{\eu^{-t} - \eu^{-2t}}{t}\dd t$.

**Solution of Exercise 9.6.**

On $\intcc{\varepsilon}{M}$, substitute $u = at$ and $u = bt$ in the two halves:

$$
\int_\varepsilon^M \frac{f(at) - f(bt)}{t}\dd t
= \int_{a\varepsilon}^{aM}\frac{f(u)}{u}\dd u
- \int_{b\varepsilon}^{bM}\frac{f(u)}{u}\dd u
= \int_{a\varepsilon}^{b\varepsilon} \frac{f(u)}{u}\dd u
- \int_{aM}^{bM} \frac{f(u)}{u}\dd u .
$$

First piece: $f(u) = f(0) + o(1)$ near $0$, and $\int_{a\varepsilon}
^{b\varepsilon} \frac{\dd u}{u} = \ln\frac ba$: the piece tends to $f(0)\ln\frac ba$. Second piece: $f(u) \to f(\infty)$, same computation: tends to $f(\infty)\ln\frac ba$. Hence the [improper integral](#def-b2-integration-improper) [converges](#def-b2-integration-improper) to $\bigl(f(0) - f(\infty)\bigr)\ln\frac ba$.

With $f(t) = \eu^{-t}$ ($f(0) = 1$, $f(\infty) = 0$), $a = 1$, $b
= 2$:

$$
\int_0^\infty \frac{\eu^{-t} - \eu^{-2t}}{t}\dd t = \ln 2 .
$$

**Exercise 9.7 ★★.**

Justify and compute $\lim_{n\to\infty} \displaystyle\int_0^n
\Bigl(1 - \frac tn\Bigr)^{\!n} t^{x-1}\,\dd t$ for $x > 0$ *(dominated convergence with $\varphi(t) = \eu^{-t}t^{x-1}$, using $(1 - t/n)^n \leq \eu^{-t}$; the limit is $\Gamma(x)$)*.

**Solution of Exercise 9.7.**

Extend the integrand by $0$ beyond $t = n$: $g_n(t) = (1 -
\frac tn)^n t^{x-1}\mathbf{1}_{t \leq n}$. Pointwise, $g_n(t) \to
\eu^{-t}t^{x-1}$ (the compound-interest limit, Year 1 volume). Domination: $\ln(1 - u) \leq -u$ gives $(1 - \frac tn)^n \leq
\eu^{-t}$ on $\intcc{0}{n}$, so $\abs{g_n(t)} \leq
\eu^{-t}t^{x-1} = \varphi(t)$, integrable. Dominated convergence:

$$
\int_0^n \Bigl(1 - \frac tn\Bigr)^n t^{x-1}\dd t
\xrightarrow[n\to\infty]{} \int_0^\infty \eu^{-t}t^{x-1}\dd t
= \Gamma(x) .
$$

(Computing the left side by repeated parts gives Euler’s product form $\Gamma(x) = \lim \frac{n!\,n^x}{x(x+1)\cdots(x+n)}$.)

**Exercise 9.8 ★★★.**

(The Gaussian integral by a parameter trick) For $x \geq 0$ set

$$
G(x) = \Bigl(\int_0^x \eu^{-t^2}\dd t\Bigr)^{\!2},
\qquad
H(x) = \int_0^1 \frac{\eu^{-x^2(1+t^2)}}{1 + t^2}\,\dd t .
$$

Prove that $G' + H' = 0$ (differentiate $H$ under the integral and substitute $u = xt$ in the resulting integral), deduce $G(x) + H(x)
= \frac\pi4$ for all $x$, and conclude

$$
\int_0^{\infty} \eu^{-t^2}\,\dd t = \frac{\sqrt\pi}{2} .
$$

**Solution of Exercise 9.8.**

$H$ is differentiable in $x$ (integrand $C^1$ in $x$, derivative $-2x(1+t^2)\cdot\frac{\eu^{-x^2(1+t^2)}}{1+t^2} =
-2x\,\eu^{-x^2}\eu^{-x^2t^2}$, [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and bounded on [compacts](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of $x$, domination over $t \in \intcc{0}{1}$ trivial):

$$
H'(x) = -2x\,\eu^{-x^2}\int_0^1 \eu^{-x^2t^2}\,\dd t
\overset{u = xt}{=} -2\,\eu^{-x^2}\int_0^x \eu^{-u^2}\,\dd u
= -G'(x),
$$

since $G'(x) = 2\eu^{-x^2}\int_0^x \eu^{-t^2}\dd t$ (chain rule on the square, fundamental theorem of calculus). So $G + H$ is constant, equal to $G(0) + H(0) = 0 + \int_0^1 \frac{\dd t}{1+t^2}
= \frac\pi4$.

As $x \to \infty$: $0 \leq H(x) \leq \eu^{-x^2}\int_0^1 \dd t \to
0$, so $G(x) \to \frac\pi4$:

$$
\int_0^\infty \eu^{-t^2}\dd t = \sqrt{\frac\pi4} =
\frac{\sqrt\pi}{2} .
$$

(Consequently $\Gamma\bigl(\frac12\bigr) = 2\int_0^\infty
\eu^{-t^2}\dd t = \sqrt\pi$, by the substitution $t = \sqrt u$.)

**Exercise 9.9 ★★★.**

Prove that $\Gamma$ is *log-convex*: $\ln\Gamma$ is convex on $\intoo{0}{\infty}$. *(Cauchy–Schwarz for integrals applied to $t^{(x+y)/2 - 1}\eu^{-t} = \bigl(t^{x-1}\eu^{-t}\bigr)^{1/2}
\bigl(t^{y-1}\eu^{-t}\bigr)^{1/2}$ gives $\Gamma\bigl(\frac{x+y}{2}\bigr)^2 \leq \Gamma(x)\Gamma(y)$; combine with [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and [Exercise 8.8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#exo-b2-realfun-8).)*

**Solution of Exercise 9.9.**

Cauchy–Schwarz (Year 1 volume, valid on $\intcc{\varepsilon}{M}$ and passed to the limit) applied to the factorization $t^{\frac{x+y}{2}-1}\eu^{-t} = \bigl(t^{x-1}\eu^{-t}\bigr)^{1/2}
\bigl(t^{y-1}\eu^{-t}\bigr)^{1/2}$:

$$
\Gamma\Bigl(\frac{x+y}{2}\Bigr) \leq
\Gamma(x)^{1/2}\,\Gamma(y)^{1/2}
\quad\Longrightarrow\quad
\ln\Gamma\Bigl(\frac{x+y}{2}\Bigr) \leq \frac{\ln\Gamma(x) +
\ln\Gamma(y)}{2} :
$$

$\ln\Gamma$ is midpoint convex; being [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 9.18](#thm-b2-integration-gammaprops)), it is convex ([Exercise 8.8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#exo-b2-realfun-8)). (Log-convexity pins $\Gamma$ down uniquely among interpolations of the factorial — the Bohr–Mollerup theorem, a Year 3 pearl.)

**Exercise 9.10 ★★★.**

(Dirichlet integral) Set $F(x) = \displaystyle\int_0^{\infty}
\frac{\sin t}{t}\,\eu^{-xt}\,\dd t$ for $x > 0$.

1. Justify $F'(x) = -\frac{1}{1 + x^2}$ (differentiate under the integral; compute $\int_0^\infty \eu^{-xt}\sin t\,\dd  t$ by two integrations by parts).
2. Prove $F(x) \to 0$ as $x \to +\infty$ and deduce $F(x) =  \frac\pi2 - \arctan x$ .
3. Admitting the [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $F$ at $0^+$ (an Abel-type theorem), conclude the value of the semi-convergent integral: $$\int_0^{\infty} \frac{\sin t}{t}\,\dd t = \frac{\pi}{2}.$$

**Solution of Exercise 9.10.**

1. On $x \geq a > 0$: the $x$-derivative of the integrand is $-\sin t\,\eu^{-xt}$, dominated by $\eu^{-at}$: [Theorem 9.11](#thm-b2-integration-leibnizrule) gives $F'(x) =  -\int_0^\infty \eu^{-xt}\sin t\,\dd t$. Two integrations by parts (or the complex exponential): $$\int_0^\infty \eu^{-xt}\sin t\,\dd t  = \Im \int_0^\infty \eu^{(-x+\iu)t}\dd t  = \Im\frac{1}{x - \iu} = \frac{1}{1 + x^2} .$$
2. $\abs{F(x)} \leq \int_0^\infty \eu^{-xt}\dd t = \frac1x \to  0$ . Integrating $F' = -\frac{1}{1+x^2}$ from $x$ to $\infty$ : $0 - F(x) = -\bigl(\frac\pi2 - \arctan x\bigr)$ , so $F(x) = \frac\pi2 - \arctan x$ .
3. Letting $x \to 0^+$ with the admitted [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) : $F(0^+) =  \frac\pi2$ , and $F(0) = \int_0^\infty \frac{\sin t}{t}\dd  t$ (the semi-convergent Dirichlet integral, [Example 9.5](#ex-b2-integration-sint) ): its value is $\frac\pi2$ .

**Exercise 9.11 ★★.**

Justify the convergence of $\displaystyle\int_0^\infty
\Bigl(\frac{\sin t}{t}\Bigr)^{\!2}\dd t$, then compute it by one integration by parts and [Exercise 9.10](#exo-b2-integration-10):

$$
\int_0^\infty \Bigl(\frac{\sin t}{t}\Bigr)^{\!2}\dd t
= \frac{\pi}{2} .
$$

(The same value as $\int_0^\infty \frac{\sin t}{t}\dd t$ — but this time the convergence is absolute.)

**Solution of Exercise 9.11.**

Convergence: near $0$ the integrand extends [continuously](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) by the value $1$ ($\sin t \sim t$); at infinity it is $\leq t^{-2}$: absolute convergence. On $\intcc{\varepsilon}{M}$, integrate by parts with $u = \sin^2 t$, $v' = t^{-2}$:

$$
\int_\varepsilon^M \frac{\sin^2 t}{t^2}\dd t
= \Bigl[-\frac{\sin^2 t}{t}\Bigr]_\varepsilon^M
+ \int_\varepsilon^M \frac{2\sin t\cos t}{t}\dd t
= \Bigl[-\frac{\sin^2 t}{t}\Bigr]_\varepsilon^M
+ \int_{2\varepsilon}^{2M} \frac{\sin u}{u}\dd u
$$

($u = 2t$ in the last integral). The bracket tends to $0$ at both ends ($\sin^2\varepsilon/\varepsilon \leq \varepsilon$; $\sin^2 M/M \leq 1/M$), and the last integral tends to $\int_0^\infty \frac{\sin u}{u}\dd u = \frac\pi2$ ([Exercise 9.10](#exo-b2-integration-10)). Hence

$$
\int_0^\infty \Bigl(\frac{\sin t}{t}\Bigr)^{\!2}\dd t
= \frac\pi2 .
$$

**Exercise 9.12 ★★★.**

(The Gaussian tail) For $x > 0$ set $T(x) = \displaystyle
\int_x^\infty \eu^{-t^2}\dd t$.

1. Writing $\eu^{-t^2} = \frac{1}{-2t}\cdot(-2t\,\eu^{-t^2})$, integrate by parts twice to obtain $$T(x) = \eu^{-x^2}\Bigl(\frac{1}{2x} -  \frac{1}{4x^3}\Bigr) +  \frac34\int_x^\infty \frac{\eu^{-t^2}}{t^4}\,\dd t .$$
2. Bound the remainder: $0 \leq \frac34\int_x^\infty  t^{-4}\eu^{-t^2}\dd t \leq \frac{3}{8x^5}\,\eu^{-x^2}$, and deduce the bracketing $$\eu^{-x^2}\Bigl(\frac{1}{2x} - \frac{1}{4x^3}\Bigr)  \leq T(x) \leq \frac{\eu^{-x^2}}{2x},  \qquad\text{hence}\qquad  T(x) \sim \frac{\eu^{-x^2}}{2x} \quad (x \to +\infty).$$
3. Why can the full alternating series obtained by iterating the parts never converge for fixed $x$ ? *(Compare the growth of the coefficients $1\cdot3\cdots(2k-1)$ with the powers $(2x^2)^k$.)*

**Solution of Exercise 9.12.**

1. Parts with $u = \frac{-1}{2t}$, $v' = -2t\,\eu^{-t^2}$ (so $v = \eu^{-t^2}$): $$T(x) = \Bigl[\frac{-\eu^{-t^2}}{2t}\Bigr]_x^\infty  - \int_x^\infty \frac{\eu^{-t^2}}{2t^2}\dd t  = \frac{\eu^{-x^2}}{2x}  - \int_x^\infty \frac{\eu^{-t^2}}{2t^2}\dd t .$$ Same device on the new integral ($u = \frac{-1}{4t^3}$, $v' = -2t\,\eu^{-t^2}$): $$\int_x^\infty \frac{\eu^{-t^2}}{2t^2}\dd t  = \frac{\eu^{-x^2}}{4x^3}  - \frac34\int_x^\infty \frac{\eu^{-t^2}}{t^4}\dd t ,$$ whence the announced identity.
2. One more integration by parts bounds the remainder: $$\int_x^\infty \frac{\eu^{-t^2}}{t^4}\dd t  = \frac{\eu^{-x^2}}{2x^5}  - \frac52\int_x^\infty\frac{\eu^{-t^2}}{t^6}\dd t  \leq \frac{\eu^{-x^2}}{2x^5},$$ so $0 \leq \frac34\int_x^\infty t^{-4}\eu^{-t^2}\dd t  \leq \frac{3}{8x^5}\eu^{-x^2}$. Dropping the (positive) remainder in the identity of question 1 gives the lower bound; dropping the (negative) second term of the first parts gives $T(x) \leq \frac{\eu^{-x^2}}{2x}$. Dividing the bracketing by $\frac{\eu^{-x^2}}{2x}$: the ratio is squeezed between $1 - \frac{1}{2x^2}$ and $1$, so $T(x)  \sim \frac{\eu^{-x^2}}{2x}$.
3. Iterating the parts produces the formal series $$T(x) \approx \frac{\eu^{-x^2}}{2x}\Bigl(1 -  \frac{1}{2x^2} + \frac{1\cdot3}{(2x^2)^2} -  \frac{1\cdot3\cdot5}{(2x^2)^3} + \cdots\Bigr),$$ whose $k$-th coefficient $1\cdot3\cdots(2k-1) =  \frac{(2k)!}{2^k k!}$ grows faster than any geometric sequence: for fixed $x$ the terms $\frac{1\cdot3\cdots(2k-1)}{(2x^2)^k}$ tend to infinity (their ratio is $\frac{2k+1}{2x^2} \to \infty$), so the series diverges for every $x$. It is an *asymptotic* expansion: truncated at any fixed order, the error is of the order of the first omitted term as $x \to \infty$ — but never a convergent series. (This tail estimate is the standard Gaussian tail bound of the probability chapters.)

## 9.5 Problem: Euler’s integrals — Beta, Gamma, and Gauss’s limit formula

**Problem 9.1.**

The $\Gamma$ function of [Definition 9.17](#def-b2-integration-gamma) is one half of Euler’s calculus of integrals; the other half is the *Beta function*

$$
B(x, y) = \int_0^1 t^{x-1}(1 - t)^{y-1}\,\dd t .
$$

This problem develops the pair $(\Gamma, B)$ with the tools of this chapter only — integration by parts, substitution, dominated convergence — and culminates in *Euler’s Beta–Gamma formula* $B(x,y) = \frac{\Gamma(x)\Gamma(y)}
{\Gamma(x+y)}$ on the half-integers and in *Gauss’s limit formula* for $\Gamma$. Along the way the Wallis integrals of [Lemma 6.11](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#lem-b2-comparison-wallis) reappear as Beta values, and Legendre’s duplication formula drops out.

**Part I — Fine structure of $\Gamma$.**

1. Recall why $\Gamma(x) = \int_0^\infty t^{x-1}\eu^{-t}\dd  t$ [converges](#def-b2-integration-improper) exactly for $x > 0$, and show $$\Gamma(x) \sim \frac1x \qquad (x \to 0^+)$$ *(functional equation plus [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $\Gamma$ at $1$)*.
2. Prove $\Gamma\bigl(\tfrac12\bigr) = \sqrt\pi$ *(substitute $t = u^2$ and invoke [Exercise 9.8](#exo-b2-integration-8))* , and deduce $\int_\R \eu^{-u^2/2}\dd u = \sqrt{2\pi}$ .
3. Show by induction, for $n \in \N$: $$\Gamma\Bigl(n + \frac12\Bigr)  = \frac{(2n)!}{4^n\,n!}\,\sqrt\pi .$$
4. Justify $\Gamma''(x) = \int_0^\infty  t^{x-1}\eu^{-t}(\ln t)^2\dd t > 0$ , and deduce that $\Gamma$ is strictly convex, attains a unique minimum at some $x_0 \in \intoo{1}{2}$ *($\Gamma(1) = \Gamma(2)  = 1$ and Rolle)* , decreases on $\intoo{0}{x_0}$ and increases on $\intoo{x_0}{\infty}$ .
5. Show that $\Gamma$ beats every power: for each $k \in  \N$ , $x^k = o\bigl(\Gamma(x)\bigr)$ as $x \to +\infty$ *(squeeze $x$ between integers and use $\Gamma(n+1)  = n!$ with the monotonicity of question 4)* .

**Part II — The [Beta function](#pb-b2-integration-1), by parts.**

6. Show that $B(x,y)$ [converges](#def-b2-integration-improper) exactly for $x > 0$ and $y >  0$ , and that $B(x,y) = B(y,x)$ .
7. Compute $B(x, 1) = \frac1x$, and prove by integration by parts, for $x, y > 0$: $$B(x, y+1) = \frac{y}{x}\,B(x+1, y) .$$
8. From the splitting $t^{x-1}(1-t)^{y-1} =  t^{x}(1-t)^{y-1} + t^{x-1}(1-t)^{y}$ deduce $B(x,y) =  B(x+1,y) + B(x,y+1)$, and combine with question 7 into the *descent relations* $$B(x, y+1) = \frac{y}{x+y}\,B(x,y),  \qquad  B(x+1, y) = \frac{x}{x+y}\,B(x,y) .$$
9. Deduce, for $m, n \geq 1$ integers: $$B(m, n) = \frac{(m-1)!\,(n-1)!}{(m+n-1)!}  = \frac{1}{(m+n-1)\binom{m+n-2}{m-1}} .$$
10. Prove *Euler’s formula for one integer argument*: for every $x > 0$ and $n \in \N^*$, $$B(x, n) = \frac{\Gamma(x)\,\Gamma(n)}{\Gamma(x + n)}$$ *(induction on $n$: both sides equal $\frac1x$ at $n  = 1$ and obey the same descent relation)*.

**Part III — Wallis integrals as Beta values.**

11. Substitute $t = \sin^2\theta$ to obtain the trigonometric form $$B(x, y) = 2\int_0^{\pi/2}  \sin^{2x-1}\theta\,\cos^{2y-1}\theta\,\dd\theta .$$
12. Deduce $W_n = \frac12\,B\bigl(\frac{n+1}2, \frac12\bigr)$ for the Wallis integral $W_n = \int_0^{\pi/2}\sin^n  \theta\,\dd\theta$ , and recover the recurrence $W_n =  \frac{n-1}{n}W_{n-2}$ of [Lemma 6.11](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#lem-b2-comparison-wallis) from the descent relations of question 8 alone.
13. Compute $B\bigl(\frac12, \frac12\bigr) = 2W_0 = \pi$ and check it against $\Gamma\bigl(\frac12\bigr)^2/\Gamma(1)$ : Euler’s formula holds at $\bigl(\frac12,  \frac12\bigr)$ .
14. Derive the closed form $W_{2n} = \frac\pi2\,  \frac{(2n)!}{4^n(n!)^2}$ from the recurrence, and verify $$B\Bigl(n + \frac12, \frac12\Bigr)  = \frac{\Gamma\bigl(n + \frac12\bigr)\Gamma\bigl(  \frac12\bigr)}{\Gamma(n+1)} .$$ Conclude, by induction with the descent relations, that Euler’s formula $B(x,y) =  \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ holds whenever $2x$ and $2y$ are positive integers.
15. Substitute $u = \frac{t}{1-t}$ to obtain the third classical form $$B(x,y) = \int_0^\infty  \frac{u^{x-1}}{(1+u)^{x+y}}\,\dd u ,$$ and check the case $x = y = \frac12$ directly ($u = v^2$ reduces it to $\int_0^\infty\frac{2\,\dd v}{1+v^2}$).

**Part IV — Gauss’s limit formula.**

16. For $x > 0$ and $n \in \N^*$, prove by $n$ successive integrations by parts: $$\int_0^n \Bigl(1 - \frac tn\Bigr)^{\!n} t^{x-1}\,\dd t  = \frac{n!\;n^x}{x(x+1)\cdots(x+n)} .$$
17. Conclude with [Exercise 9.7](#exo-b2-integration-7) (dominated convergence) *Gauss’s limit formula*: $$\Gamma(x) = \lim_{n\to\infty}  \frac{n!\;n^x}{x(x+1)\cdots(x+n)}  \qquad (x > 0).$$
18. Taking logarithms, show that for $x > 0$: $$\ln\Gamma(x) = -\ln x - \gamma x +  \sum_{k=1}^{\infty}\Bigl(\frac xk -  \ln\Bigl(1 + \frac xk\Bigr)\Bigr),$$ where $\gamma$ is Euler’s constant ([Example 6.7](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-harmonic)); justify the convergence of the series *(the general term is $\sim \frac{x^2}{2k^2}$)*.
19. Use Gauss’s formula at $x = \frac12$ and the central binomial asymptotics $\binom{2n}{n} \sim  \frac{4^n}{\sqrt{\pi n}}$ ( [Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial) ) to recompute $\Gamma\bigl(\frac12\bigr) = \sqrt\pi$ : Stirling’s constant and the Gaussian integral are the same number in two disguises.
20. Check that Gauss’s formula reproves the functional equation: from the exact identity $$\frac{n!\,n^{x+1}}{(x+1)\cdots(x+n+1)}  = \frac{n!\,n^{x}}{x(x+1)\cdots(x+n)}\cdot  \frac{n\,x}{x+n+1},$$ conclude $\Gamma(x+1) = x\,\Gamma(x)$ again. (Gauss’s formula determines $\Gamma$ outright; the Year 3 volume proves the sharper Bohr–Mollerup theorem: the functional equation plus log-convexity already pin $\Gamma$ down.)

**Part V — Dividends.**

21. For $a > 0$ show $\int_0^\infty \eu^{-t^a}\dd t =  \Gamma\bigl(1 + \frac1a\bigr)$ , and compute the limit as $a \to +\infty$ by dominated convergence *(pointwise limit $\mathbf 1_{t < 1}$; dominate by $1$ on $\intoc{0}{1}$ and by $\eu^{-t^2}$ beyond, for $a \geq  2$)* . Check the answer against the [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $\Gamma$ .
22. For $n \geq 1$ show $$\int_0^1 \frac{\dd t}{\sqrt{1 - t^n}}  = \frac1n\,B\Bigl(\frac1n, \frac12\Bigr),$$ and recover the values $2$ ($n = 1$) and $\frac\pi2$ ($n  = 2$). (For $n = 4$ this is the *lemniscate constant*, which has no elementary closed form; its story belongs to the theory of elliptic integrals.)
23. (Moments) For $x > 0$ and $k \in \N$, show $$\frac{1}{\Gamma(x)}\int_0^\infty  t^{k}\,t^{x-1}\eu^{-t}\,\dd t  = \frac{\Gamma(x+k)}{\Gamma(x)}  = x(x+1)\cdots(x+k-1),$$ the rising factorial; check that $x = 1$ gives $k!$. (In the probability chapters this is the $k$-th moment of a standard waiting-time density.)
24. Prove the Beta identity, valid for *all* $x > 0$: $$B(x, x) = 2^{1-2x}\,B\Bigl(x, \frac12\Bigr)$$ *(substitute $t = \frac{1+s}2$, exploit the symmetry in $s$, then set $s = \sqrt v$)*. Deduce, for $2x \in \N^*$, *Legendre’s duplication formula* $$\Gamma(x)\,\Gamma\Bigl(x + \frac12\Bigr)  = 2^{1-2x}\,\sqrt\pi\;\Gamma(2x),$$ and verify it directly at $x = n$ via question 3. (For general $x$ it follows from the same identity once Euler’s formula is known for all arguments — the double-integral proof in the chapter on multiple integrals.)
25. Synthesis. In one sentence each: (i) where integration by parts carried the whole of Part II; (ii) where dominated convergence entered Parts IV and V; (iii) which asymptotic inputs were imported from the comparison chapter; (iv) what is now proved of Euler’s formula $B(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x+y)$ , and what remains for the double integral to settle.

**Solution of Problem 9.1.**

**1.** At $0^+$ the integrand is $\sim t^{x-1}$: the finite-endpoint scale [converges](#def-b2-integration-improper) iff $1 - x < 1$, i.e. $x > 0$ (and for $x \leq 0$, $t^{x-1} \geq t^{-1}$ diverges); at $+\infty$, $t^{x-1}\eu^{-t} = o(t^{-2})$ [converges](#def-b2-integration-improper) for every $x$. Then $\Gamma(x) = \frac{\Gamma(x+1)}{x}$ and $\Gamma(x+1)
\to \Gamma(1) = 1$ as $x \to 0^+$ ([continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), [Theorem 9.18](#thm-b2-integration-gammaprops)): $\Gamma(x) \sim
\frac1x$.

**2.** With $t = u^2$, $\dd t = 2u\,\dd u$:

$$
\Gamma\Bigl(\frac12\Bigr) = \int_0^\infty
t^{-1/2}\eu^{-t}\dd t
= \int_0^\infty \frac{\eu^{-u^2}}{u}\,2u\,\dd u
= 2\int_0^\infty \eu^{-u^2}\dd u = \sqrt\pi
$$

by [Exercise 9.8](#exo-b2-integration-8). With $u = v/\sqrt2$:

$$
\int_\R \eu^{-v^2/2}\dd v
= 2\sqrt2\int_0^\infty \eu^{-u^2}\dd u
= \sqrt2\,\sqrt\pi = \sqrt{2\pi} .
$$

**3.** True for $n = 0$ (both sides $\sqrt\pi$). If $\Gamma(n + \frac12) = \frac{(2n)!}{4^n n!}\sqrt\pi$, the functional equation gives

$$
\Gamma\Bigl(n + 1 + \frac12\Bigr)
= \Bigl(n + \frac12\Bigr)\Gamma\Bigl(n + \frac12\Bigr)
= \frac{2n+1}{2}\cdot\frac{(2n)!}{4^n n!}\sqrt\pi
= \frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt\pi ,
$$

the last step because $\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$ and $\frac{2n+1}{2} = \frac{(2n+2)(2n+1)}{4(n+1)}$.

**4.** [Theorem 9.18](#thm-b2-integration-gammaprops) gives $\Gamma''(x) = \int_0^\infty t^{x-1}\eu^{-t}(\ln t)^2\dd t$ (two applications of the Leibniz rule, dominations as in the theorem’s proof); the integrand is $\geq 0$ and not identically zero, so $\Gamma'' > 0$: $\Gamma$ is strictly convex and $\Gamma'$ is strictly increasing. Since $\Gamma(1) = \Gamma(2)
= 1$, Rolle provides $x_0 \in \intoo12$ with $\Gamma'(x_0) =
0$; strict monotonicity of $\Gamma'$ makes $x_0$ its unique zero, with $\Gamma' < 0$ before and $\Gamma' > 0$ after: $\Gamma$ decreases on $\intoo0{x_0}$, increases on $\intoo{x_0}\infty$, and $x_0$ is the unique minimum.

**5.** Let $k \in \N$ and $x \geq 3$; choose the integer $n$ with $n + 1 \leq x < n + 2$ (so $n \geq 1$). By the monotonicity of question 4 (valid from $x_0 < 2$ on): $\Gamma(x)
\geq \Gamma(n + 1) = n!$, while $x^k \leq (n+2)^k$. Since $\frac{n!}{(n+2)^k} \to \infty$ (factorials beat powers, Year 1 volume), $\frac{\Gamma(x)}{x^k} \geq \frac{n!}{(n+2)^k} \to
\infty$ as $x \to \infty$: $x^k = o(\Gamma(x))$.

**6.** Near $0$ the integrand is $\sim t^{x-1}$ (convergent iff $x > 0$), near $1$ it is $\sim (1-t)^{y-1}$ (iff $y > 0$); both comparisons are between positive functions, so $B(x,y)$ [converges](#def-b2-integration-improper) exactly for $x, y > 0$. The substitution $t \mapsto 1 - t$ swaps the two factors: $B(x,y) = B(y,x)$.

**7.** $B(x,1) = \int_0^1 t^{x-1}\dd t = \frac1x$. Parts on $\intcc\varepsilon{1-\varepsilon}$ with $u = (1-t)^y$, $v =
\frac{t^x}{x}$:

$$
\int t^{x-1}(1-t)^{y}\dd t
= \Bigl[\frac{t^x(1-t)^y}{x}\Bigr]
+ \frac{y}{x}\int t^{x}(1-t)^{y-1}\dd t ;
$$

the bracket vanishes at both ends as $\varepsilon \to 0$ ($x >
0$ at $0$, $y > 0$ at $1$), leaving $B(x, y+1) = \frac
yx\,B(x+1, y)$.

**8.** Since $t + (1-t) = 1$:

$$
t^{x-1}(1-t)^{y-1} = t^{x}(1-t)^{y-1} + t^{x-1}(1-t)^{y},
$$

so $B(x,y) = B(x+1,y) + B(x,y+1)$. Question 7 reads $B(x+1,y) =
\frac xy B(x,y+1)$; substituting,

$$
B(x,y) = \Bigl(\frac xy + 1\Bigr)B(x,y+1)
= \frac{x+y}{y}\,B(x,y+1),
$$

i.e. $B(x,y+1) = \frac{y}{x+y}B(x,y)$; the twin relation follows by the symmetry of question 6.

**9.** Induction on $n$ at fixed $m$: $B(m,1) = \frac1m =
\frac{(m-1)!\,0!}{m!}$, and if the formula holds at $n$,

$$
B(m, n+1) = \frac{n}{m+n}\,B(m,n)
= \frac{n}{m+n}\cdot\frac{(m-1)!(n-1)!}{(m+n-1)!}
= \frac{(m-1)!\,n!}{(m+n)!} .
$$

Rewriting: $B(m,n) = \frac{(m-1)!(n-1)!}{(m+n-1)!} =
\bigl[(m+n-1)\binom{m+n-2}{m-1}\bigr]^{-1}$.

**10.** Both sides of $B(x,n) =
\frac{\Gamma(x)\Gamma(n)}{\Gamma(x+n)}$ equal $\frac1x$ at $n =
1$ ($\Gamma(1) = 1$, $\Gamma(x+1) = x\Gamma(x)$). If they agree at $n$, then by the descent relation and the functional equation:

$$
B(x, n+1) = \frac{n}{x+n}\,B(x,n), \qquad
\frac{\Gamma(x)\Gamma(n+1)}{\Gamma(x+n+1)}
= \frac{n}{x+n}\cdot
\frac{\Gamma(x)\Gamma(n)}{\Gamma(x+n)} :
$$

the two sequences obey the same recursion from the same seed, hence agree for all $n \in \N^*$ and all $x > 0$.

**11.** With $t = \sin^2\theta$ ($\theta \in
\intoo0{\pi/2}$, $\dd t = 2\sin\theta\cos\theta\,\dd\theta$), $t^{x-1} = \sin^{2x-2}\theta$ and $(1-t)^{y-1} =
\cos^{2y-2}\theta$:

$$
B(x,y) = \int_0^{\pi/2}\sin^{2x-2}\theta\,\cos^{2y-2}\theta
\cdot 2\sin\theta\cos\theta\,\dd\theta
= 2\int_0^{\pi/2}\sin^{2x-1}\theta\,\cos^{2y-1}\theta\,
\dd\theta .
$$

**12.** Take $y = \frac12$ (killing the cosine factor) and $2x - 1 = n$: $B\bigl(\frac{n+1}2, \frac12\bigr) = 2W_n$, i.e. $W_n = \frac12 B\bigl(\frac{n+1}2,\frac12\bigr)$. The descent relation in the first variable gives

$$
\frac{W_n}{W_{n-2}}
= \frac{B\bigl(\frac{n-1}2 + 1, \frac12\bigr)}
{B\bigl(\frac{n-1}2, \frac12\bigr)}
= \frac{\frac{n-1}2}{\frac{n-1}2 + \frac12}
= \frac{n-1}{n} :
$$

the Wallis recurrence, this time with no integration by parts on sines — Part II did the work once and for all.

**13.** $B\bigl(\frac12,\frac12\bigr) = 2W_0 =
2\cdot\frac\pi2 = \pi$, while $\Gamma\bigl(\frac12\bigr)^2/\Gamma(1) = (\sqrt\pi)^2 = \pi$: Euler’s formula holds at $\bigl(\frac12,\frac12\bigr)$.

**14.** Iterating $W_{2n} = \frac{2n-1}{2n}W_{2n-2}$ from $W_0 = \frac\pi2$:

$$
W_{2n} = \frac\pi2\prod_{k=1}^{n}\frac{2k-1}{2k}
= \frac\pi2\cdot\frac{(2n)!}{4^n(n!)^2},
$$

since $\prod(2k-1) = \frac{(2n)!}{2^n n!}$ and $\prod 2k = 2^n
n!$. Hence, using question 3:

$$
B\Bigl(n+\frac12, \frac12\Bigr) = 2W_{2n}
= \pi\,\frac{(2n)!}{4^n(n!)^2}
= \frac{(2n)!\sqrt\pi}{4^n n!}\cdot\frac{\sqrt\pi}{n!}
= \frac{\Gamma\bigl(n+\frac12\bigr)\Gamma\bigl(\frac12\bigr)}
{\Gamma(n+1)} .
$$

Now fix $x \in \frac12\N^*$. Euler’s formula holds at $(x,
\frac12)$: for $x$ integer this is question 10 (with symmetry), for $x = n + \frac12$ it is the display above. Both sides of Euler’s formula obey the descent recursion $y \mapsto y + 1$ (question 8 on the left, the functional equation on the right, as in question 10): induction propagates the formula from $y =
\frac12$ and $y = 1$ to every $y \in \frac12\N^*$. Euler’s formula therefore holds whenever $2x, 2y \in \N^*$.

**15.** With $u = \frac{t}{1-t}$, i.e. $t =
\frac{u}{1+u}$, $1 - t = \frac{1}{1+u}$, $\dd t =
\frac{\dd u}{(1+u)^2}$:

$$
B(x,y) = \int_0^\infty
\Bigl(\frac{u}{1+u}\Bigr)^{x-1}
\Bigl(\frac{1}{1+u}\Bigr)^{y-1}
\frac{\dd u}{(1+u)^2}
= \int_0^\infty \frac{u^{x-1}}{(1+u)^{x+y}}\,\dd u .
$$

At $x = y = \frac12$, with $u = v^2$:

$$
\int_0^\infty \frac{u^{-1/2}}{1+u}\dd u
= \int_0^\infty \frac{2\,\dd v}{1+v^2} = \pi
= B\Bigl(\frac12,\frac12\Bigr) . \checkmark
$$

**16.** One integration by parts, for $1 \leq k \leq n$ and $s > 0$ ($u = (1 - t/n)^k$, $v = t^s/s$; the boundary terms vanish):

$$
\int_0^n \Bigl(1-\frac tn\Bigr)^{\!k} t^{s-1}\dd t
= \frac{k}{ns}\int_0^n \Bigl(1-\frac tn\Bigr)^{\!k-1}
t^{s}\dd t .
$$

Starting from $k = n$, $s = x$ and iterating $n$ times:

$$
\int_0^n \Bigl(1-\frac tn\Bigr)^{\!n} t^{x-1}\dd t
= \frac{n(n-1)\cdots1}{n^n\,x(x+1)\cdots(x+n-1)}
\int_0^n t^{x+n-1}\dd t
= \frac{n!}{n^n}\cdot
\frac{n^{x+n}}{x(x+1)\cdots(x+n)} ,
$$

which is $\dfrac{n!\,n^x}{x(x+1)\cdots(x+n)}$.

**17.** By [Exercise 9.7](#exo-b2-integration-7) the left side tends to $\Gamma(x)$ (dominated convergence with dominator $t^{x-1}\eu^{-t}$); the right side is Gauss’s quotient:

$$
\Gamma(x) = \lim_{n\to\infty}
\frac{n!\,n^x}{x(x+1)\cdots(x+n)} .
$$

**18.** Taking logarithms in question 16’s quotient $G_n(x)$ and splitting $\ln(x+k) = \ln k + \ln(1 + x/k)$ for $k
\geq 1$:

$$
\ln G_n(x) = x\ln n - \ln x - \sum_{k=1}^n
\ln\Bigl(1+\frac xk\Bigr)
= -\ln x + x(\ln n - H_n)
+ \sum_{k=1}^n\Bigl(\frac xk -
\ln\Bigl(1+\frac xk\Bigr)\Bigr).
$$

For $u \geq 0$, $u - \frac{u^2}2 \leq \ln(1+u) \leq u$, so the general term lies in $\intcc{0}{x^2/(2k^2)}$: the series [converges](#def-b2-integration-improper) (comparison with $\sum k^{-2}$). Since $\ln n - H_n
\to -\gamma$ ([Example 6.7](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-harmonic)) and $\ln G_n(x)
\to \ln\Gamma(x)$ (question 17 and [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $\ln$):

$$
\ln\Gamma(x) = -\ln x - \gamma x
+ \sum_{k=1}^\infty\Bigl(\frac xk -
\ln\Bigl(1+\frac xk\Bigr)\Bigr) .
$$

**19.** At $x = \frac12$, the denominator is $\prod_{k=0}^n\bigl(k+\frac12\bigr) =
\frac{(2n+1)!}{2^{2n+1}n!}$ (multiply out the halves), so

$$
G_n\Bigl(\frac12\Bigr)
= \frac{n!\,\sqrt n\;2^{2n+1}n!}{(2n+1)!}
= \frac{2\sqrt n\;4^n}{(2n+1)\binom{2n}{n}} .
$$

With $\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$ ([Example 6.14](https://one-course.com/books/math/4/en/chapter/6-comparison-of-functions#ex-b2-comparison-centralbinomial)):

$$
G_n\Bigl(\frac12\Bigr)
\sim \frac{2\sqrt n\,\sqrt{\pi n}}{2n+1}
\longrightarrow \sqrt\pi = \Gamma\Bigl(\frac12\Bigr) .
$$

The central binomial coefficient’s $\sqrt\pi$ (which came from Wallis, hence from Stirling’s constant) and the Gaussian integral’s $\sqrt\pi$ are the same number.

**20.** The identity is direct algebra: multiply $\frac{n!\,n^x}{x(x+1)\cdots(x+n)}$ by $\frac{nx}{x+n+1}$ and absorb $x$ into the product, $n$ into $n^x$. Letting $n \to
\infty$: the left side tends to $\Gamma(x+1)$ (Gauss at $x+1$), the right side to $\Gamma(x)\cdot x\cdot 1$ since $\frac{n}{x+n+1} \to 1$: $\Gamma(x+1) = x\Gamma(x)$ — recovered without a single integration by parts.

**21.** With $u = t^a$, $t = u^{1/a}$, $\dd t = \frac1a
u^{1/a - 1}\dd u$:

$$
\int_0^\infty \eu^{-t^a}\dd t
= \frac1a\int_0^\infty u^{\frac1a - 1}\eu^{-u}\dd u
= \frac1a\,\Gamma\Bigl(\frac1a\Bigr)
= \Gamma\Bigl(1 + \frac1a\Bigr) .
$$

As $a \to +\infty$ (along any sequence): $\eu^{-t^a} \to 1$ for $0 < t < 1$, $\to \eu^{-1}$ at $t = 1$, $\to 0$ for $t > 1$; for $a \geq 2$ dominate by $\mathbf 1_{t \leq 1} +
\eu^{-t^2}\mathbf 1_{t > 1}$ ($t^a \geq t^2$ for $t \geq 1$), integrable. Dominated convergence: the integral tends to $\int_0^1 1\,\dd t = 1$ — as it must, since $\Gamma(1 +
\frac1a) \to \Gamma(1) = 1$ by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity).

**22.** With $u = t^n$, $\dd t = \frac1n u^{1/n - 1}\dd
u$:

$$
\int_0^1 \frac{\dd t}{\sqrt{1-t^n}}
= \frac1n\int_0^1 u^{\frac1n-1}(1-u)^{-1/2}\dd u
= \frac1n\,B\Bigl(\frac1n, \frac12\Bigr) .
$$

$n = 1$: $B\bigl(1,\frac12\bigr) = B\bigl(\frac12,1\bigr) = 2$, matching $\int_0^1\frac{\dd t}{\sqrt{1-t}} = 2$. $n = 2$: $\frac12 B\bigl(\frac12,\frac12\bigr) = \frac\pi2 = \arcsin 1$. For $n = 4$ the value $\frac14 B\bigl(\frac14,\frac12\bigr)$ is the lemniscate constant: no elementary closed form.

**23.** Iterating the functional equation:

$$
\frac{1}{\Gamma(x)}\int_0^\infty t^{x+k-1}\eu^{-t}\dd t
= \frac{\Gamma(x+k)}{\Gamma(x)}
= (x+k-1)(x+k-2)\cdots x ,
$$

the rising factorial with $k$ factors. At $x = 1$: $\Gamma(1+k)/\Gamma(1) = k!$, the moments of $\eu^{-t}$ from [Exercise 9.2](#exo-b2-integration-2).

**24.** Substitute $t = \frac{1+s}2$ ($s \in
\intoo{-1}1$, $\dd t = \frac{\dd s}2$, $t(1-t) =
\frac{1-s^2}4$):

$$
B(x,x) = \int_{-1}^{1}\Bigl(\frac{1-s^2}{4}\Bigr)^{x-1}
\frac{\dd s}{2}
= 4^{1-x}\int_0^1 (1-s^2)^{x-1}\dd s
$$

(the integrand is even). Then $s = \sqrt v$ ($\dd s =
\frac{\dd v}{2\sqrt v}$):

$$
B(x,x) = \frac{4^{1-x}}{2}\int_0^1
v^{-1/2}(1-v)^{x-1}\dd v
= 2^{1-2x}\,B\Bigl(\frac12, x\Bigr) .
$$

For $2x \in \N^*$ every argument in sight lies in $\frac12\N^*$, so Euler’s formula (question 14) applies to both sides:

$$
\frac{\Gamma(x)^2}{\Gamma(2x)}
= 2^{1-2x}\,
\frac{\Gamma\bigl(\frac12\bigr)\Gamma(x)}
{\Gamma\bigl(x+\frac12\bigr)}
\quad\Longleftrightarrow\quad
\Gamma(x)\,\Gamma\Bigl(x+\frac12\Bigr)
= 2^{1-2x}\sqrt\pi\;\Gamma(2x) .
$$

Direct check at $x = n$: the left side is $(n-1)!\cdot
\frac{(2n)!\sqrt\pi}{4^n n!} = \frac{(2n)!\sqrt\pi}{4^n n}$, the right side $2\cdot4^{-n}\sqrt\pi\,(2n-1)! =
\frac{(2n)!\sqrt\pi}{4^n n}$: equal.

**25.** (i) Integration by parts produced $B(x,y+1) =
\frac yx B(x+1,y)$, the single identity from which every descent relation, the integer and half-integer values, and the Wallis recurrence all flow. (ii) Dominated convergence turned the elementary integrals $\int_0^n(1-t/n)^n t^{x-1}$ into $\Gamma(x)$ (Gauss’s formula, question 17) and computed the limit $a \to \infty$ in question 21. (iii) From the comparison chapter we imported Euler’s constant ($\ln n - H_n \to
-\gamma$, question 18) and the central binomial asymptotics (question 19) — i.e. Stirling’s formula in disguise. (iv) Euler’s formula $B(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x+y)$ is now proved for $y \in \N^*$ with $x > 0$ arbitrary (question 10) and for all half-integer pairs (question 14); the general case $x, y > 0$ awaits the double-integral computation of the chapter on multiple integrals, which factorizes $\Gamma(x)\Gamma(y)$ over a quarter-plane.