University Mathematics — Year 2 · Bachelor Year 2

9Integration

The Year 1 volume built the integral on a segment. This chapter extends it to arbitrary intervals (improper integrals, with the full comparison toolkit), then studies integrals depending on a parameter — 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 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 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 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, 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 on $\intco{a}{b}$ :

$\int_a^b f$ converges iff the primitive $x \mapsto \int_a^x f$ is bounded;
$f \leq g$ : convergence of $\int g$ forces that of $\int f$ ; divergence transfers the other way;
$f \sim g$ at $b$ : the two integrals have the same nature;
the reference scales: at $+\infty$ , $\int^{\infty} \frac{\dd t}{t^\alpha}$ converges 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 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, 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: 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 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.

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 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). 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): $I = \frac\pi2$ . 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) 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: 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 absolutely ( $\abs{\cos t}/t^2 \leq t^{-2}$ ). But not absolutely: 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 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 without converging absolutely — 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 on an interval $I$ , converging pointwise to a piecewise continuous $f$ , 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, 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 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). 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) 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 (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, $I$ an interval) with: $t \mapsto f(x, t)$ piecewise continuous for each $x$ ; $x \mapsto f(x, t)$ continuous 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 on $A$ .

Proof. Definedness: domination gives absolute convergence. 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 in $x$ ) under the fixed domination $\varphi$ : dominated convergence gives $F(x_n) \to F(x_0)$ ; conclude by the sequential characterization of continuity (Definition 4.5). ∎

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 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 by Theorem 9.10 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 (Theorem 9.10). 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. 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 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 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 $\int_a^b f$ :

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.
If $f$ has constant sign near that end, find an equivalent and compare with the reference scales of Theorem 9.2.
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; minorations like $\abs{\sin t} \geq \sin^2t$ detect genuine semi-convergence.
For a value, not just the nature: parts, substitution, or a parameter (differentiate a simpler integral, as in Example 9.12 and Example 9.21).
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 half-line, since 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 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.

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 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” 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 and differentiation under the integral sign produce the Fourier coefficients’ calculus in the Fourier chapter, where every $c_n(f)$ is a parameter integral 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 $\Gamma$ 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 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. $\Gamma(1) = \int \eu^{-t} = 1$ ; induction gives the factorial.

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 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; 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: 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 $\Gamma$ goes from here)

The weekend problem of this chapter builds the whole Euler calculus around $\Gamma$ : the Beta function, 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 convergent, dominated by $\eu^{-t^2}$ ). By Theorem 9.11 (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 — and again, by double integration, in Chapter 20.

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

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); near $1$ , $\sim (1-t)^{-1/2}$ : converges. 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

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

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 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

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 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

Solution of Exercise 9.5.

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.

$C^2$ 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. 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 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 at $0$ and the limit at $\infty$ .) Compute $\int_0^\infty \frac{\eu^{-t} - \eu^{-2t}}{t}\dd t$ .

Solution

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 converges 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

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

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 and bounded on compacts 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 and Exercise 8.8.)

Solution

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 (Theorem 9.18), it is convex (Exercise 8.8). (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 $x > 0$ .

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).
Prove $F(x) \to 0$ as $x \to +\infty$ and deduce $F(x) = \frac\pi2 - \arctan x$ .
Admitting the 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

Solution of Exercise 9.10.

On $x \geq a > 0$ : the $x$ -derivative of the integrand is $-\sin t\,\eu^{-xt}$ , dominated by $\eu^{-at}$ : Theorem 9.11 gives $F'(x) = -\int_0^\infty \eu^{-xt}\sin t\,\dd t$ . Two integrations by parts (or the complex exponential):

$\int_0^\infty \eu^{-xt}\sin t\,\dd t = \Im \int_0^\infty \eu^{(-x+\iu)t}\dd t = \Im\frac{1}{x - \iu} = \frac{1}{1 + x^2} .$
$\abs{F(x)} \leq \int_0^\infty \eu^{-xt}\dd t = \frac1x \to 0$ . Integrating $F' = -\frac{1}{1+x^2}$ from $x$ to $\infty$ : $0 - F(x) = -\bigl(\frac\pi2 - \arctan x\bigr)$ , so $F(x) = \frac\pi2 - \arctan x$ .
Letting $x \to 0^+$ with the admitted 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): 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:

\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

Solution of Exercise 9.11.

Convergence: near $0$ the integrand extends continuously 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). 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 $T(x) = \displaystyle \int_x^\infty \eu^{-t^2}\dd t$ .

Writing $\eu^{-t^2} = \frac{1}{-2t}\cdot(-2t\,\eu^{-t^2})$ , integrate by parts twice to obtain

$T(x) = \eu^{-x^2}\Bigl(\frac{1}{2x} - \frac{1}{4x^3}\Bigr) + \frac34\int_x^\infty \frac{\eu^{-t^2}}{t^4}\,\dd t .$
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

$\eu^{-x^2}\Bigl(\frac{1}{2x} - \frac{1}{4x^3}\Bigr) \leq T(x) \leq \frac{\eu^{-x^2}}{2x}, \qquad\text{hence}\qquad T(x) \sim \frac{\eu^{-x^2}}{2x} \quad (x \to +\infty).$
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

Solution of Exercise 9.12.

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.
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 $T(x) \sim \frac{\eu^{-x^2}}{2x}$ .
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 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 reappear as Beta values, and Legendre’s duplication formula drops out.

Part I — Fine structure of $\Gamma$ .

Recall why $\Gamma(x) = \int_0^\infty t^{x-1}\eu^{-t}\dd t$ converges exactly for $x > 0$ , and show

$\Gamma(x) \sim \frac1x \qquad (x \to 0^+)$

(functional equation plus continuity of $\Gamma$ at $1$ ).
Prove $\Gamma\bigl(\tfrac12\bigr) = \sqrt\pi$ (substitute $t = u^2$ and invoke Exercise 9.8), and deduce $\int_\R \eu^{-u^2/2}\dd u = \sqrt{2\pi}$ .
Show by induction, for $n \in \N$ :

$\Gamma\Bigl(n + \frac12\Bigr) = \frac{(2n)!}{4^n\,n!}\,\sqrt\pi .$
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 $\intoo{x_0}{\infty}$ .
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, by parts.

Show that $B(x,y)$ converges exactly for $x > 0$ and $y > 0$ , and that $B(x,y) = B(y,x)$ .
Compute $B(x, 1) = \frac1x$ , and prove by integration by parts, for $x, y > 0$ :

$B(x, y+1) = \frac{y}{x}\,B(x+1, y) .$
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

$B(x, y+1) = \frac{y}{x+y}\,B(x,y), \qquad B(x+1, y) = \frac{x}{x+y}\,B(x,y) .$
Deduce, for $m, n \geq 1$ integers:

$B(m, n) = \frac{(m-1)!\,(n-1)!}{(m+n-1)!} = \frac{1}{(m+n-1)\binom{m+n-2}{m-1}} .$
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.

Substitute $t = \sin^2\theta$ to obtain the trigonometric form

$B(x, y) = 2\int_0^{\pi/2} \sin^{2x-1}\theta\,\cos^{2y-1}\theta\,\dd\theta .$
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 from the descent relations of question 8 alone.
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 $\bigl(\frac12, \frac12\bigr)$ .
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.
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.

For $x > 0$ and $n \in \N^*$ , prove by $n$ successive integrations by parts:

$\int_0^n \Bigl(1 - \frac tn\Bigr)^{\!n} t^{x-1}\,\dd t = \frac{n!\;n^x}{x(x+1)\cdots(x+n)} .$
Conclude with Exercise 9.7 (dominated convergence) Gauss’s limit formula:

$\Gamma(x) = \lim_{n\to\infty} \frac{n!\;n^x}{x(x+1)\cdots(x+n)} \qquad (x > 0).$
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); justify the convergence of the series (the general term is $\sim \frac{x^2}{2k^2}$ ).
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) to recompute $\Gamma\bigl(\frac12\bigr) = \sqrt\pi$ : Stirling’s constant and the Gaussian integral are the same number in two disguises.
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.

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 of $\Gamma$ .
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.)
(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.)
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.)
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

Solution of Problem 9.1.

1. At $0^+$ the integrand is $\sim t^{x-1}$ : the finite-endpoint scale converges 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 for every $x$ . Then $\Gamma(x) = \frac{\Gamma(x+1)}{x}$ and $\Gamma(x+1) \to \Gamma(1) = 1$ as $x \to 0^+$ (continuity, Theorem 9.18): $\Gamma(x) \sim \frac1x$ .

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. 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 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 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 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 (comparison with $\sum k^{-2}$ ). Since $\ln n - H_n \to -\gamma$ (Example 6.7) and $\ln G_n(x) \to \ln\Gamma(x)$ (question 17 and 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):

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.

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.

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.