---
title: "The Lebesgue Integral"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 10
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral
---

# Chapter 10 — The Lebesgue Integral

Riemann’s integral slices the *domain* into small intervals; Lebesgue’s slices the *range*: to integrate $f$, [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) the sets $\{f > t\}$. The change looks innocent and is revolutionary. Limits and integrals, forever quarreling in the Riemann theory (uniform convergence required!), are reconciled by three convergence theorems — monotone convergence, Fatou, dominated convergence — whose hypotheses are almost embarrassingly weak. This chapter constructs the integral over an arbitrary [measure space](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $(X, \mathcal A,
\mu)$, proves the three theorems, settles the exact relationship with Riemann’s integral (a bounded function is [Riemann-integrable](#def-b3-lebesgue-l1) iff it is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) [almost everywhere](#def-b3-lebesgue-l1)), and industrializes the differentiation of parameter-dependent integrals — the technique that the weekend problem uses to compute $\int_0^\infty\frac{\sin x}x\,\dd x$ and $\int_\R \eu^{-x^2}\dd x$.

## 10.1 Measurable functions

**Definition 10.1.**

Let $(X, \mathcal A)$, $(Y, \mathcal B)$ be measurable spaces. $f \colon X \to Y$ is *measurable* if $f^{-1}(B) \in \mathcal A$ for every $B \in
\mathcal B$. For real (or $[-\infty,+\infty]$-valued) functions, $Y = \R$ carries its Borel $\sigma$-algebra, and it suffices to check $f^{-1}(\intoo t{+\infty}) = \{f > t\} \in
\mathcal A$ for all $t \in \R$: the good sets $\{B :
f^{-1}(B) \in \mathcal A\}$ form a $\sigma$-algebra (preimages commute with set operations) containing the generating rays ([Definition 9.2](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-borel), [Method 9.17](https://one-course.com/books/math/5/en/chapter/9-measure-theory#met-b3-measure-goodsets)).

**Proposition 10.2.**

(a) Compositions of [measurable](#def-b3-lebesgue-measurable) maps are [measurable](#def-b3-lebesgue-measurable); [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) are [Borel-measurable](#def-b3-lebesgue-measurable). (b) If $f, g \colon X \to \R$ are [measurable](#def-b3-lebesgue-measurable), so are $f + g$, $fg$, $\max(f,g)$, $\abs f$, $\lambda f$. (c) If $(f_n)$ are [measurable](#def-b3-lebesgue-measurable) with values in $[-\infty, +\infty]$, then $\sup_nf_n$, $\inf_nf_n$, $\limsup f_n$, $\liminf f_n$ are [measurable](#def-b3-lebesgue-measurable); if $f_n \to f$ pointwise, $f$ is [measurable](#def-b3-lebesgue-measurable).

**Proof.** (a) $(g\circ f)^{-1}(B) = f^{-1}(g^{-1}(B))$; [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) gives [measurability](#def-b3-lebesgue-measurable) via the generating [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) ([Problem 9.1](https://one-course.com/books/math/5/en/chapter/9-measure-theory#pb-b3-measure-1), question 10, in general form). (b) $(f, g) \colon X \to \R^2$ is [measurable](#def-b3-lebesgue-measurable) for the Borel $\sigma$-algebra of $\R^2$ — check on open boxes, which generate ($\R^2$’s opens are countable unions of rational boxes): $(f,g)^{-1}(U\times V) = f^{-1}(U)\cap g^{-1}(V)$ — and $+, \times, \max$ are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\R^2 \to \R$: compose. (c) $\{\sup f_n > t\} = \bigcup_n\{f_n > t\}$; $\inf = -\sup(-)$; $\limsup = \inf_N\sup_{n \geq N}$; a pointwise limit is its own $\limsup$. ∎

**Definition 10.3.**

A *simple function* is a [measurable function](#def-b3-lebesgue-measurable) with finitely many values: $s = \sum_{i=1}^n
c_i\,\mathbf 1_{A_i}$, $A_i \in \mathcal A$ disjoint, $c_i \geq
0$ (for the nonnegative theory). Its integral is

$$
\int s\,\dd\mu = \sum_i c_i\,\mu(A_i) \in [0, +\infty]
$$

(convention $0\cdot\infty = 0$); the value does not depend on the [representation](https://one-course.com/books/math/5/en/chapter/5-representations-of-finite-groups#def-b3-representations-rep) (refine two partitions).

**Theorem 10.4 (Approximation by simple functions).**

Every [measurable](#def-b3-lebesgue-measurable) $f \colon X \to [0, +\infty]$ is the pointwise limit of an *increasing* sequence of [simple functions](#def-b3-lebesgue-simple):

$$
s_n = \sum_{k=1}^{n2^n} \frac{k-1}{2^n}\,
\mathbf 1_{\{\frac{k-1}{2^n} \leq f < \frac k{2^n}\}}
+ n\,\mathbf 1_{\{f \geq n\}} \nearrow f .
$$

**Proof.** Each $s_n$ is simple (the sets are preimages of Borel sets). Monotonicity: passing from $n$ to $n+1$ splits each dyadic level in two and never decreases the assigned value (a point with $\frac{k-1}{2^n} \leq f(x) < \frac k{2^n}$ gets either $\frac{2k-2}{2^{n+1}}$ or $\frac{2k-1}{2^{n+1}}$, both $\geq
\frac{k-1}{2^n}$; the cap $n$ rises too). Convergence: if $f(x) < \infty$, for $n > f(x)$ we have $f(x) - s_n(x) \leq
2^{-n}$; if $f(x) = \infty$, $s_n(x) = n \to \infty$. ∎

## 10.2 The integral and the convergence theorems

**Definition 10.5.**

For [measurable](#def-b3-lebesgue-measurable) $f \geq 0$:

$$
\int f \,\dd\mu = \sup\Bigl\{\int s\,\dd\mu : s \text{ simple},
\ 0 \leq s \leq f\Bigr\} \in [0, +\infty].
$$

It is monotone in $f$ by construction, and extends the simple case (for simple $f$, the sup is attained at $f$: comparison of simple integrals via common refinements).

**Theorem 10.6 (Monotone convergence, Beppo Levi).**

If $0 \leq f_n \nearrow f$ pointwise ([measurable](#def-b3-lebesgue-measurable)), then

$$
\int f_n\,\dd\mu \nearrow \int f\,\dd\mu .
$$

**Proof.** $f$ is [measurable](#def-b3-lebesgue-measurable) ([Proposition 10.2](#prop-b3-lebesgue-stability)(c)) and $\int f_n$ increases to some $L \leq \int f$ (monotonicity). Conversely, fix a simple $s = \sum c_i\mathbf 1_{A_i} \leq f$ and $\theta \in (0,1)$; the sets $E_n = \{f_n \geq \theta s\}$ are [measurable](#def-b3-lebesgue-measurable) and increase to $X$ (where $s(x) > 0$: $f(x)
\geq s(x) > \theta s(x)$, so eventually $f_n(x) \geq \theta
s(x)$; where $s(x) = 0$: trivially). Then

$$
\int f_n \geq \int_{E_n}\theta s\,\dd\mu
= \theta\sum_i c_i\,\mu(A_i \cap E_n)
\xrightarrow[n\to\infty]{} \theta\sum_ic_i\,\mu(A_i)
= \theta\int s
$$

by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from below ([Proposition 9.6](https://one-course.com/books/math/5/en/chapter/9-measure-theory#prop-b3-measure-basics)(c)). So $L \geq \theta\int s$ for all $\theta < 1$ and all simple $s
\leq f$: $L \geq \int f$. ∎

**Corollary 10.7.**

For [measurable](#def-b3-lebesgue-measurable) $f, g \geq 0$ and $c \geq 0$: $\int(f + g) =
\int f + \int g$ and $\int cf = c\int f$; for a series of nonnegative [measurable functions](#def-b3-lebesgue-measurable), $\int\sum_nf_n =
\sum_n\int f_n$.

**Proof.** For [simple functions](#def-b3-lebesgue-simple), additivity is a computation on a common refinement. In general take $s_n \nearrow f$, $t_n \nearrow g$ ([Theorem 10.4](#thm-b3-lebesgue-approximation)): $s_n + t_n \nearrow f +
g$, and MCT passes additivity to the limit. The series statement is MCT applied to the partial sums. ∎

**Theorem 10.8 (Fatou’s lemma).**

For [measurable](#def-b3-lebesgue-measurable) $f_n \geq 0$:

$$
\int \liminf_n f_n \,\dd\mu \;\leq\; \liminf_n \int
f_n\,\dd\mu .
$$

**Proof.** Let $g_N = \inf_{n\geq N}f_n$: [measurable](#def-b3-lebesgue-measurable), $0 \leq g_N \nearrow
\liminf f_n$, and $g_N \leq f_n$ for every $n \geq N$, so $\int g_N \leq \inf_{n \geq N}\int f_n$. Apply MCT to the left side: $\int\liminf f_n = \lim_N\int g_N \leq
\lim_N\inf_{n\geq N}\int f_n = \liminf\int f_n$. ∎

**Definition 10.9.**

A [measurable](#def-b3-lebesgue-measurable) $f \colon X \to \R$ (or $\C$) is *integrable* if $\int\abs
f\,\dd\mu < \infty$; then $\int f = \int f^+ - \int f^-$ (positive and negative parts; real and imaginary parts in the complex case). The integral is linear on integrable functions (decompose and recombine positive parts; the complex case reduces to the real one) and satisfies $\abs{\int f} \leq
\int\abs f$ (real case: $\pm\int f = \int(\pm f) \leq
\int\abs f$; complex case: multiply by a unimodular constant to make the integral real). A property holds *almost everywhere* (a.e.) if it fails only on a $\mu$-null set; modifying $f$ on a null set changes no integral (the difference is dominated by $\infty\cdot\mathbf
1_N$, of integral $0$).

**Theorem 10.10 (Dominated convergence).**

Let $f_n \to f$ a.e., with $\abs{f_n} \leq g$ a.e. for a fixed *[integrable](#def-b3-lebesgue-l1)* $g$. Then $f$ is [integrable](#def-b3-lebesgue-l1) and

$$
\int f_n\,\dd\mu \longrightarrow \int f\,\dd\mu,
\qquad\text{indeed}\quad \int\abs{f_n - f}\,\dd\mu \to 0 .
$$

**Proof.** Discard a null set to make the hypotheses pointwise. $\abs f
\leq g$: $f$ is [integrable](#def-b3-lebesgue-l1). The functions $h_n = 2g - \abs{f_n
- f} \geq 0$ satisfy $\liminf h_n = 2g$; Fatou gives

$$
\int 2g \leq \liminf\int\bigl(2g - \abs{f_n - f}\bigr)
= \int 2g - \limsup\int\abs{f_n - f},
$$

so $\limsup\int\abs{f_n - f} \leq 0$ (the subtraction is legal: $\int 2g < \infty$). Finally $\abs{\int f_n - \int f} \leq
\int\abs{f_n - f} \to 0$. ∎

**Method 10.11.**

Faced with $\lim_n\int f_n$: try, in order — (1) is the sequence monotone (or a series of nonnegative terms)? MCT, no [integrability](#def-b3-lebesgue-l1) needed. (2) Is there a single [integrable](#def-b3-lebesgue-l1) dominator $g \geq \abs{f_n}$, found by crude bounds (“$\sup_n$” the estimates)? DCT. (3) No domination, no monotonicity? Fatou still bounds one side, and equality may genuinely fail: the escaping bump $f_n = n\mathbf
1_{\intoo0{1/n}}$ has $\int f_n = 1$ but $f_n \to 0$ a.e. Domination is exactly what forbids mass from escaping to infinity, vertically or horizontally.

## 10.3 Riemann versus Lebesgue

**Theorem 10.12 (Lebesgue’s criterion).**

Let $f \colon \intcc ab \to \R$ be *bounded*. Then $f$ is [Riemann-integrable](#def-b3-lebesgue-l1) iff $f$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\lambda$-almost everywhere; in that case $f$ is [Lebesgue-integrable](#def-b3-lebesgue-l1) and the two integrals coincide.

**Proof.** For a subdivision $\sigma = (a = x_0 < \dots < x_N = b)$, let $L_\sigma$ and $U_\sigma$ be the step functions equal, on each $\intoo{x_{i-1}}{x_i}$, to $m_i = \inf_{[x_{i-1}, x_i]}f$ and $M_i = \sup$; the Darboux sums are their integrals (Riemann and Lebesgue agree on step functions, both giving $\sum
m_i\Delta x_i$). Take a sequence of subdivisions $\sigma_n$, each refining the last, of mesh $\to 0$, with Darboux sums converging to the lower and upper Darboux integrals of $f$. The refinements make $L_{\sigma_n}$ nondecreasing and $U_{\sigma_n}$ nonincreasing pointwise off the countable set $D$ of all division points; call the limits $\ell$ and $u$ ([measurable](#def-b3-lebesgue-measurable), [Proposition 10.2](#prop-b3-lebesgue-stability)). For $x \notin
D$, writing $I_n(x)$ for the open $\sigma_n$-interval containing $x$: $\ell(x) = \sup_n\inf_{I_n(x)}f$ and $u(x) =
\inf_n\sup_{I_n(x)}f$; since the meshes shrink to $0$, these are the lower and upper *envelopes* of $f$ at $x$ — $u(x) - \ell(x)$ is the oscillation of $f$ at $x$ — so that *$\ell(x) = u(x)$ iff $f$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $x$*. By MCT/DCT (bounded, finite interval):

$$
\int_{\intcc ab}\ell\,\dd\lambda = \lim_n\int L_{\sigma_n}
= \underline{\int}f,
\qquad
\int_{\intcc ab}u\,\dd\lambda = \overline{\int}f .
$$

$f$ [Riemann-integrable](#def-b3-lebesgue-l1) $\iff$ $\underline\int f =
\overline\int f$ $\iff$ $\int(u - \ell) = 0$ $\iff$ $u = \ell$ a.e. ($u - \ell \geq 0$; [Exercise 10.5](#exo-b3-lebesgue-5)) $\iff$ $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) a.e. In that case $\ell \leq f \leq u$ with $\ell =
u$ a.e.: $f$ equals the [measurable](#def-b3-lebesgue-measurable) $\ell$ a.e., hence is [Lebesgue-measurable](#def-b3-lebesgue-measurable) ([completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) of $\lambda$) with $\int f\,\dd\lambda = \int\ell\,\dd\lambda = \underline\int f =
\int_a^bf$. ∎

**Example 10.13.**

$\mathbf 1_\Q$ is nowhere [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): not [Riemann-integrable](#def-b3-lebesgue-l1) — but Lebesgue-trivial: $\int\mathbf 1_\Q\,\dd\lambda =
\lambda(\Q) = 0$. Thomae’s function ( $\frac1q$ at rationals $\frac pq$, $0$ elsewhere) is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) exactly at the irrationals: [Riemann-integrable](#def-b3-lebesgue-l1) with integral $0$. And *improper* Riemann integrals are a different notion: $\int_0^\infty\frac{\sin x}x\,\dd x$ converges as a limit of $\int_0^A$ (the weekend problem computes it $= \frac\pi2$), but $\frac{\sin x}x \notin L^1(\intoo0{+\infty})$: the absolute integral diverges like the harmonic series ([Exercise 10.6](#exo-b3-lebesgue-6)). Lebesgue’s theory trades conditional convergence for robust limit theorems.

## 10.4 Integrals with parameters

Throughout, $(X, \mathcal A, \mu)$ is a [measure space](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure), $T$ a metric space (the parameter), and $f \colon T \times X \to \C$ with $f(t, \cdot)$ [integrable](#def-b3-lebesgue-l1) for each $t$; set $F(t) = \int_X
f(t, x)\,\dd\mu(x)$.

**Theorem 10.14 (Continuity).**

Suppose: $t \mapsto f(t,x)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $t_0$ for a.e. $x$, and there is an [integrable](#def-b3-lebesgue-l1) $g$ with $\abs{f(t,x)} \leq
g(x)$ for all $t$ in a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $t_0$ and a.e. $x$. Then $F$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $t_0$.

**Proof.** For any sequence $t_n \to t_0$: $f(t_n, \cdot) \to f(t_0,
\cdot)$ a.e., dominated by $g$: DCT gives $F(t_n) \to F(t_0)$; sequential [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) suffices in metric spaces ([Remark 6.8](https://one-course.com/books/math/5/en/chapter/6-general-topology#rem-b3-topology-sequences)). ∎

**Theorem 10.15 (Differentiation under the integral).**

Let $T$ be an open interval of $\R$. Suppose: for a.e. $x$, $t \mapsto f(t,x)$ is differentiable on $T$, with

$$
\Bigl|\frac{\partial f}{\partial t}(t, x)\Bigr| \leq g(x)
\quad \text{for all } t \in T,\ \text{a.e. } x,
$$

$g$ [integrable](#def-b3-lebesgue-l1). Then $F$ is differentiable on $T$ with $F'(t)
= \int_X \frac{\partial f}{\partial t}(t, x)\,\dd\mu(x)$.

**Proof.** Fix $t$ and $h_n \to 0$: the difference quotients

$$
\varphi_n(x) = \frac{f(t + h_n, x) - f(t, x)}{h_n}
\longrightarrow \frac{\partial f}{\partial t}(t,x)
\quad\text{a.e.},
$$

and the mean value inequality bounds $\abs{\varphi_n(x)} \leq
\sup_{s}\abs{\partial_tf(s,x)} \leq g(x)$: DCT applies, and $\frac{F(t + h_n) - F(t)}{h_n} = \int\varphi_n \to
\int\partial_t f(t, \cdot)$. ∎

**Example 10.16 (The Gamma function).**

For $t > 0$ let

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

The integral converges: near $0$, $x^{t-1}$ is [integrable](#def-b3-lebesgue-l1) ($t >
0$); at infinity, $x^{t-1}\eu^{-x} \leq C\eu^{-x/2}$. Integration by parts (on $[\varepsilon, A]$, then limits via MCT) gives the functional equation $\Gamma(t + 1) =
t\,\Gamma(t)$, whence $\Gamma(n+1) = n!$: the factorial interpolated. On every $\intcc ab \subseteq \intoo0{+\infty}$, $\partial_t\bigl(x^{t-1}\eu^{-x}\bigr) = \ln
x\cdot x^{t-1}\eu^{-x}$ is dominated by $\abs{\ln x}(x^{a-1} +
x^{b-1})\eu^{-x}$, [integrable](#def-b3-lebesgue-l1): $\Gamma$ is $\mathcal C^1$, and by induction $\mathcal C^\infty$, with $\Gamma^{(k)}(t) =
\int_0^\infty(\ln x)^kx^{t-1}\eu^{-x}\dd x$. The value $\Gamma(\frac12) = \sqrt\pi$ is the Gaussian integral in disguise ([Problem 10.1](#pb-b3-lebesgue-1)).

![The integrand xx: the improper integral ∈t_0∈fty converges by alternating cancellation between the arches, but the areas | | of the arches behave like 2π k — a harmonic series: xx ∉ L1. Lebesgue integrability is absolute integrability.](https://one-course.com/images/onecourse/chapters/math-5/b3-lebesgue/fig-8a5288e0e9f0.svg)

*The integrand $\frac{\sin x}x$: the improper integral $\int_0^\infty$ converges by alternating cancellation between the arches, but the areas $\abs{\cdot}$ of the arches behave like $\frac2{\pi k}$ — a harmonic series: $\frac{\sin x}x
\notin L^1$. Lebesgue [integrability](#def-b3-lebesgue-l1) is absolute [integrability](#def-b3-lebesgue-l1).*

## 10.5 Exercises

**Exercise 10.1 ★.**

(a) Show that a monotone function $\R \to \R$ is [Borel-measurable](#def-b3-lebesgue-measurable), and that a derivative (of an everywhere differentiable function) is [Borel-measurable](#def-b3-lebesgue-measurable). (b) Show that $f \colon X \to \R$ is [measurable](#def-b3-lebesgue-measurable) iff $\{f > q\}
\in \mathcal A$ for every *rational* $q$.

**Solution of Exercise 10.1.**

(a) If $f$ is nondecreasing, $\{f > t\}$ is $\varnothing$, $\R$, or a ray $\intoo a{+\infty}$ / $\intco a{+\infty}$: Borel in every case; nonincreasing likewise. A derivative: $f'(x) =
\lim_n n\bigl(f(x + \frac1n) - f(x)\bigr)$ is a pointwise limit of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (hence [measurable](#def-b3-lebesgue-measurable)) functions: [Proposition 10.2](#prop-b3-lebesgue-stability)(c).

(b) $\{f > t\} = \bigcup_{q \in \Q,\, q > t}\{f > q\}$: if the rational levels are [measurable](#def-b3-lebesgue-measurable), all levels are, and the rays generate $\mathcal B(\R)$.

**Exercise 10.2 ★.**

Compute, with full justification:

$$
\lim_{n\to\infty}\int_0^{+\infty}
\frac{\cos x}{(1 + x/n)^{n}}\,\dd x,
\qquad
\lim_{n\to\infty}\int_0^1 \frac{n\,x^{n-1}}{1 + x}\,\dd x .
$$

*(For the second: substitute $u = x^n$ before dominating.)*

**Solution of Exercise 10.2.**

First: $(1 + x/n)^n \nearrow \eu^x$ for $x \geq 0$, so the integrand tends pointwise to $\eu^{-x}\cos x$; for $n \geq 2$, $(1 + x/n)^n \geq (1 + x/2)^2$, giving the [integrable](#def-b3-lebesgue-l1) dominator $(1 + x/2)^{-2}$. DCT:

$$
\lim_n\int_0^\infty\frac{\cos x}{(1 + x/n)^n}\dd x
= \int_0^\infty \eu^{-x}\cos x\,\dd x
= \operatorname{Re}\int_0^\infty\eu^{-(1 - \iu)x}\dd x
= \operatorname{Re}\frac{1}{1 - \iu} = \frac12 .
$$

Second: substitute $u = x^n$ (a $\mathcal C^1$ bijection of $\intoo01$):

$$
\int_0^1\frac{nx^{n-1}}{1 + x}\dd x = \int_0^1\frac{\dd u}{1 +
u^{1/n}} \longrightarrow \int_0^1\frac{\dd u}{2} = \frac12,
$$

by DCT: for $u \in \intoo01$, $u^{1/n} \to 1$, and the integrand is bounded by $1$ on a finite [measure space](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure).

**Exercise 10.3 ★★.**

(a) Exhibit strict inequality in Fatou’s lemma. (b) Exhibit $f_n \to 0$ pointwise with $\int f_n = 1$ in three ways: escape in height, in width, to infinity. Which single hypothesis of DCT does each violate? (c) Show that in Fatou’s lemma one cannot replace $\liminf$ by $\limsup$ on either side.

**Solution of Exercise 10.3.**

(a) $f_n = n\,\mathbf 1_{\intoo0{1/n}}$: $\liminf f_n = 0$ pointwise, $\int f_n = 1$: $0 < 1$.

(b) Height: $n\mathbf 1_{\intoo0{1/n}}$; width: $\frac1n\mathbf
1_{\intoo0n}$; translation: $\mathbf 1_{\intoo n{n+1}}$. All tend to $0$ pointwise with $\int = 1$. In each case the *domination* hypothesis fails: $\sup_nf_n$ is $\approx 1/x$ near $0$, $\approx$ a nonintegrable constant profile, $\mathbf 1_{\intoo1\infty}$-like — never [integrable](#def-b3-lebesgue-l1).

(c) “$\int\limsup f_n \geq \limsup\int f_n$” fails for the translating bump: left side $0$, right side $1$. “$\limsup\int \leq \int\limsup$” is the same statement. And Fatou for $\limsup$ with $\leq$ reversed (“reverse Fatou”) requires a dominator — the same bump is the counterexample.

**Exercise 10.4 ★★.**

(a) Show $\displaystyle\int_0^{+\infty}\frac{x}{\eu^x -
1}\,\dd x = \sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}6$ *(expand $\frac1{\eu^x - 1}$ in a geometric series and integrate term by term — which theorem permits it?)*. (b) (Sophomore’s dream) Show $\displaystyle\int_0^1 x^{-x}\,\dd x = \sum_{n\geq1}n^{-n}$. *(Write $x^{-x} = \eu^{-x\ln x} = \sum_k\frac{(-x\ln
x)^k}{k!}$ and compute $\int_0^1(-x\ln x)^k\dd x$ by substituting $x = \eu^{-u/(k+1)}$, recognizing $\Gamma$.)*

**Solution of Exercise 10.4.**

(a) For $x > 0$: $\frac1{\eu^x - 1} = \frac{\eu^{-x}}{1 -
\eu^{-x}} = \sum_{n\geq1}\eu^{-nx}$, so $\frac{x}{\eu^x - 1} =
\sum_{n\geq1}x\eu^{-nx}$, a series of nonnegative [measurable functions](#def-b3-lebesgue-measurable): [Corollary 10.7](#cor-b3-lebesgue-additivity) allows term-by-term integration:

$$
\int_0^\infty\frac{x\,\dd x}{\eu^x - 1}
= \sum_{n\geq1}\int_0^\infty x\eu^{-nx}\dd x
= \sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}6
$$

($\int_0^\infty x\eu^{-nx}\dd x = n^{-2}$ by parts; Basel from the Year 2 volume, or [Exercise 13.5](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#exo-b3-hilbert-5) to come).

(b) On $\intoo01$, $-x\ln x \geq 0$, so $x^{-x} =
\eu^{-x\ln x} = \sum_k\frac{(-x\ln x)^k}{k!}$ is a series of nonnegative terms: interchange again. Substituting $x =
\eu^{-u/(k+1)}$:

$$
\int_0^1(-x\ln x)^k\dd x
= \int_0^\infty\Bigl(\frac{u}{k+1}\Bigr)^{k}
\eu^{-\frac{ku}{k+1}}\;\frac{\eu^{-\frac u{k+1}}}{k+1}\,\dd u
= \frac{1}{(k+1)^{k+1}}\int_0^\infty u^k\eu^{-u}\dd u
= \frac{k!}{(k+1)^{k+1}} .
$$

Hence $\int_0^1x^{-x}\dd x = \sum_{k\geq0}\frac{1}{(k+1)^{k+1}}
= \sum_{n\geq1}n^{-n}$: the sophomore’s dream, rigorously.

**Exercise 10.5 ★★.**

(a) Show that $f \geq 0$ [measurable](#def-b3-lebesgue-measurable) with $\int f\,\dd\mu = 0$ satisfies $f = 0$ a.e. *(Consider $\{f \geq 1/n\}$ and Markov’s inequality: $\mu(\{f \geq a\}) \leq \frac1a\int f$ — prove it.)* (b) Show that an [integrable](#def-b3-lebesgue-l1) $f$ is finite a.e. (c) Show that if $\int_A f\,\dd\mu = 0$ for *every* [measurable](#def-b3-lebesgue-measurable) $A$, then $f = 0$ a.e.

**Solution of Exercise 10.5.**

(a) Markov: $a\,\mathbf 1_{\{f \geq a\}} \leq f$, integrate: $\mu(\{f \geq a\}) \leq \frac1a\int f$. If $\int f = 0$: $\mu(\{f \geq \frac1n\}) = 0$ for every $n$, and $\{f > 0\} =
\bigcup_n\{f \geq \frac1n\}$ is null.

(b) $\mu(\{\abs f = \infty\}) \leq \mu(\{\abs f \geq n\}) \leq
\frac1n\int\abs f \to 0$.

(c) Take $A = \{f > 0\}$: $\int f^+\dd\mu = \int_Af\,\dd\mu =
0$, so $f^+ = 0$ a.e. by (a); likewise $f^- = 0$ a.e.

**Exercise 10.6 ★★.**

(a) Apply [Theorem 10.12](#thm-b3-lebesgue-riemann) to decide Riemann [integrability](#def-b3-lebesgue-l1) of: $\mathbf 1_\Q$; Thomae’s function; $\mathbf
1_K$ for $K$ a fat Cantor set ([Exercise 9.5](https://one-course.com/books/math/5/en/chapter/9-measure-theory#exo-b3-measure-5)). (b) Show that $\int_1^{+\infty}\abs{\frac{\sin x}x}\,\dd x =
+\infty$, while $\lim_{A\to\infty}\int_1^A\frac{\sin
x}x\,\dd x$ exists *(integrate by parts)*: improper convergence without [integrability](#def-b3-lebesgue-l1).

**Solution of Exercise 10.6.**

(a) $\mathbf 1_\Q$: discontinuous everywhere, not [Riemann-integrable](#def-b3-lebesgue-l1) ([Theorem 10.12](#thm-b3-lebesgue-riemann)); its Lebesgue integral is $\lambda(\Q) = 0$. Thomae: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at every irrational (given $\varepsilon$, only finitely many rationals in $\intcc01$ have denominator $\leq
1/\varepsilon$; avoid them by a small [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)), discontinuous at rationals ([density](#ex-b3-lebesgue-gamma) of irrationals): [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) a.e., [Riemann-integrable](#def-b3-lebesgue-l1), integral $0$ (it vanishes a.e.). $\mathbf 1_K$, $K$ a fat Cantor set: the discontinuity set is $\partial K = K$ (closed with empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior)), of [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) $\frac12 > 0$: *not* [Riemann-integrable](#def-b3-lebesgue-l1) — yet [Lebesgue-integrable](#def-b3-lebesgue-l1) with integral $\lambda(K) = \frac12$.

(b) $\int_{k\pi}^{(k+1)\pi}\frac{\abs{\sin x}}x\dd x \geq
\frac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}\abs{\sin x}\dd x =
\frac{2}{(k+1)\pi}$: the series diverges. Convergence of the improper integral: for $A > \pi$,

$$
\int_\pi^A\frac{\sin x}x\dd x = \Bigl[\frac{-\cos
x}x\Bigr]_\pi^A - \int_\pi^A\frac{\cos x}{x^2}\dd x,
$$

and both terms converge as $A \to \infty$ ($\frac1{x^2}$ is [integrable](#def-b3-lebesgue-l1)): conditional convergence without absolute [integrability](#def-b3-lebesgue-l1).

**Exercise 10.7 ★★.**

Justify that $F(t) = \int_0^{+\infty}\eu^{-x^2}\cos(tx)\,\dd x$ is $\mathcal C^1$ on $\R$ and satisfies $F'(t) = -\frac t2F(t)$ *(integrate by parts)*; deduce $F(t) =
F(0)\,\eu^{-t^2/4}$. (With $F(0) = \frac{\sqrt\pi}2$ from the weekend problem: the Gaussian is essentially its own Fourier transform — [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) will systematize this.)

**Solution of Exercise 10.7.**

Domination: $\abs{\partial_t(\eu^{-x^2}\cos(tx))} =
\abs{x\eu^{-x^2}\sin(tx)} \leq x\eu^{-x^2}$, [integrable](#def-b3-lebesgue-l1) and independent of $t$: [Theorem 10.15](#thm-b3-lebesgue-paramdiff) applies globally,

$$
F'(t) = -\int_0^\infty x\eu^{-x^2}\sin(tx)\,\dd x
= \Bigl[\tfrac12\eu^{-x^2}\sin(tx)\Bigr]_0^\infty
- \frac t2\int_0^\infty\eu^{-x^2}\cos(tx)\dd x
= -\frac t2F(t)
$$

(integration by parts with $\dd v = x\eu^{-x^2}\dd x$). The linear ODE gives $F(t) = F(0)\eu^{-t^2/4}$; with $F(0) =
\frac{\sqrt\pi}2$ ([Problem 10.1](#pb-b3-lebesgue-1)), the Gaussian reproduces itself under this cosine transform.

**Exercise 10.8 ★★★.**

(Frullani) Let $0 < a < b$. Show

$$
\int_0^{+\infty}\frac{\eu^{-ax} - \eu^{-bx}}{x}\,\dd x =
\ln\frac ba,
$$

by writing the integrand as $\int_a^b \eu^{-xt}\,\dd t$ and justifying the interchange via the nonnegative theory ([Corollary 10.7](#cor-b3-lebesgue-additivity) in [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) form — anticipate Tonelli, or slice $[a,b]$ into $n$ equal parts and pass to the limit).

**Solution of Exercise 10.8.**

The integral converges: near $0$ the integrand tends to $b - a$ (bounded), and it decays like $\eu^{-ax}$ at infinity. Fix $a$ and view $I(b) = \int_0^\infty\frac{\eu^{-ax} -
\eu^{-bx}}x\dd x$ as a function of $b \in \intco a{+\infty}$. For all $b \geq a$: $\abs{\partial_b(\text{integrand})} =
\eu^{-bx} \leq \eu^{-ax}$, and $\int_0^\infty\eu^{-ax}\dd x =
\frac1a < \infty$: an [integrable](#def-b3-lebesgue-l1) dominator. So [Theorem 10.15](#thm-b3-lebesgue-paramdiff) gives $I'(b) =
\int_0^\infty\eu^{-bx}\dd x = \frac1b$, and $I(a) = 0$:

$$
I(b) = \int_a^b\frac{\dd t}t = \ln\frac ba .
$$

(Equivalently, the hint’s route: the integrand is $\int_a^b\eu^{-xt}\dd t \geq 0$ and the interchange is the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) analogue of [Corollary 10.7](#cor-b3-lebesgue-additivity), i.e. Tonelli — proved in [Chapter 11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ch-b3-product); the parameter route stays within this chapter.)

**Exercise 10.9 ★★.**

Let $f \geq 0$ be [measurable](#def-b3-lebesgue-measurable) on $(X, \mathcal A, \mu)$. Show that $\nu(A) = \int_A f\,\dd\mu$ defines a [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) (*density* $f$ with respect to $\mu$), and that $\int g\,\dd\nu = \int gf\,\dd\mu$ for all [measurable](#def-b3-lebesgue-measurable) $g \geq 0$ *(prove it for indicators, then [simple functions](#def-b3-lebesgue-simple), then MCT — the standard machine)*.

**Solution of Exercise 10.9.**

$\nu(\varnothing) = 0$; for disjoint $(A_n)$, $f\mathbf
1_{\bigsqcup A_n} = \sum_nf\mathbf 1_{A_n}$ (pointwise, all terms $\geq 0$), and [Corollary 10.7](#cor-b3-lebesgue-additivity) gives $\sigma$-additivity. The formula $\int g\,\dd\nu = \int
gf\,\dd\mu$: for $g = \mathbf 1_A$ it is the definition of $\nu$; for simple $g$, linearity; for $g \geq 0$ [measurable](#def-b3-lebesgue-measurable), take simple $s_n \nearrow g$ ([Theorem 10.4](#thm-b3-lebesgue-approximation)): $s_nf \nearrow gf$, and MCT on both sides passes to the limit. (This “indicator $\to$ simple $\to$ MCT” escalator is the standard machine of the theory.)

**Exercise 10.10 ★★★.**

(A Weierstrass-style failure) Define $f(t) =
\int_0^{+\infty}\frac{\sin(tx)}{x(1 + x^2)}\,\dd x$. (a) Show that $f$ is well defined and [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\R$, and $\mathcal C^1$ with $f'(t) = \int_0^\infty\frac{\cos(tx)}{1 +
x^2}\dd x$ for every $t$ — but that differentiating *again* under the integral is illegitimate. (b) Admitting $f'(t) = \frac\pi2\eu^{-t}$ for $t > 0$ (proved in [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues)), what is $\int_0^\infty\frac{x\sin(tx)}{1+x^2}\dd x$ for $t > 0$, and why does its formula confirm the failure in (a)?

**Solution of Exercise 10.10.**

(a) $\abs{\sin(tx)} \leq \abs tx$ gives $\abs{\frac{\sin(tx)}{x(1+x^2)}} \leq \frac{\abs
t}{1+x^2}$: the integral converges, and on $\abs t \leq T$ the dominator $\frac{T}{1+x^2}$ yields [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ([Theorem 10.14](#thm-b3-lebesgue-paramcont)). Differentiation: $\abs{\partial_t} = \abs{\frac{\cos(tx)}{1+x^2}} \leq
\frac1{1+x^2}$, [integrable](#def-b3-lebesgue-l1): $f'(t) =
\int_0^\infty\frac{\cos(tx)}{1+x^2}\dd x$ for all $t$. A second differentiation would require integrating $\frac{x\sin(tx)}{1 +
x^2}$, whose absolute value behaves like $\frac{\abs{\sin(tx)}}
x$ at infinity: *not* [integrable](#def-b3-lebesgue-l1) — no dominator exists and [Theorem 10.15](#thm-b3-lebesgue-paramdiff) cannot be applied again.

(b) Admitting $f'(t) = \frac\pi2\eu^{-t}$ for $t > 0$: by oddness of $f$, $f'$ is even, so $f'(t) =
\frac\pi2\eu^{-\abs t}$ — which is *not differentiable at $0$*: $f$ is $\mathcal C^1$ but not $\mathcal C^2$, confirming that the blocked second differentiation was not a technical accident. For $t > 0$ the improper integral $\int_0^\infty\frac{x\sin(tx)}{1+x^2}\dd x$ equals $-f''(t) =
\frac\pi2\eu^{-t}$ (differentiating the admitted formula where it is legitimate, i.e. on $\intoo0\infty$) — an improper, non-Lebesgue value.

**Exercise 10.11 ★★.**

(Scheffé’s lemma) Let $f_n, f \geq 0$ be [integrable](#def-b3-lebesgue-l1) with $f_n \to f$ a.e. and $\int f_n \to \int f$. (a) Show that $\int\abs{f_n - f} \to 0$. *(Apply dominated convergence to $g_n = (f - f_n)^+ \leq f$, and write $\int\abs{f_n - f} = 2\int g_n - \int(f - f_n)$.)* (b) Show by example that the hypothesis $\int f_n \to \int
f$ cannot be dropped (a sliding or concentrating bump), and that the conclusion fails for signed $f_n$ without absolute-value control: $f_n = n\mathbf 1_{\intoc0{1/n}} -
n\mathbf 1_{\intoc{-1/n}0}$ has $f_n \to 0$ a.e., $\int f_n
= 0 \to 0$, yet $\int\abs{f_n} = 2$. (c) Application (densities): if probability densities $p_n
\to p$ a.e., then automatically $\int\abs{p_n - p} \to 0$: pointwise convergence of densities is $L^1$ convergence — a convergence upgrade for free.

**Solution of Exercise 10.11.**

(a) Let $g_n = (f - f_n)^+$: then $0 \leq g_n \leq f$ (positivity of $f_n$), $g_n \to 0$ a.e., and $f$ is an [integrable](#def-b3-lebesgue-l1) dominator: $\int g_n \to 0$ (DCT). Since $\abs{f_n - f} = 2(f - f_n)^+ - (f - f_n)$,

$$
\int\abs{f_n - f} = 2\int g_n - \Bigl(\int f - \int
f_n\Bigr) \longrightarrow 0 + 0 .
$$

(b) The sliding bump $f_n = \mathbf 1_{\intcc n{n+1}}$ has $f_n \to 0$ a.e. and $\int f_n = 1 \not\to 0$: without the convergence of integrals, $L^1$ convergence fails (and so does the hypothesis). The signed example: $f_n \to 0$ at every $x \neq 0$, $\int f_n = 0$, but $\int\abs{f_n} = 2$: for signed sequences the theorem is genuinely about $\abs{f_n}$-type control, and positivity was used exactly in $g_n \leq f$.

(c) Densities satisfy $\int p_n = 1 = \int p$: the hypothesis of (a) is automatic, so $p_n \to p$ a.e. forces $\norm{p_n - p}_{L^1} \to 0$ — and hence convergence of the probabilities $\int_Ap_n \to \int_Ap$ *uniformly* over all [measurable](#def-b3-lebesgue-measurable) $A$ ($\abs{\int_A(p_n - p)} \leq
\norm{p_n - p}_1$): Scheffé turns pointwise convergence of densities into total-variation convergence of laws.

**Exercise 10.12 ★★.**

Classical limits, with full justification via MCT/DCT:

$$
\text{(a)}\ \lim_{n\to\infty}\int_0^n\Bigl(1 -
\frac xn\Bigr)^n\eu^{x/2}\,\dd x, \qquad
\text{(b)}\ \lim_{n\to\infty}\int_0^1\frac{n\,x^{n-1}}{1 +
x}\,\dd x,
$$

$$
\text{(c)}\ \lim_{n\to\infty}\int_0^\infty
\frac{\dd x}{(1 + x/n)^n\,x^{1/n}} .
$$

*(For (a): $(1 - x/n)^n \nearrow \eu^{-x}$ for fixed $x$ — prove the monotonicity via $\log$; for (b), integrate by parts or substitute $x = u^{1/n}$ and identify a boundary concentration; for (c), find an [integrable](#def-b3-lebesgue-l1) dominator valid for all $n \geq 2$ by splitting at $x =
1$.)*

**Solution of Exercise 10.12.**

(a) On $\intoo0n$, $\varphi_n(x) = n\log(1 - \frac xn)$ increases in $n$ to $-x$ (the map $t \mapsto
\frac{\log(1 - xt)}{t}$ decreases as $t = \frac1n
\downarrow 0$; or expand: $\varphi_{n}' \geq 0$ in $n$ via $\log(1-u) + \frac{u}{1-u} \geq 0$). So $(1 -
\frac xn)^n\eu^{x/2}\mathbf 1_{x<n} \nearrow
\eu^{-x/2}$, and MCT gives

$$
\lim_n\int_0^n\Bigl(1 - \frac xn\Bigr)^n\eu^{x/2}\dd x =
\int_0^\infty\eu^{-x/2}\dd x = 2 .
$$

(b) Substitute $u = x^n$ ($x = u^{1/n}$, $n x^{n-1}\dd x =
\dd u$):

$$
\int_0^1\frac{nx^{n-1}}{1 + x}\dd x =
\int_0^1\frac{\dd u}{1 + u^{1/n}} .
$$

For $u \in \intoo01$: $u^{1/n} \to 1$, so the integrand tends to $\frac12$, dominated by $1$: the limit is $\frac12$ (DCT). (The mass of $nx^{n-1}$ concentrates at $x
= 1$, where $\frac1{1+x} = \frac12$: the substitution makes the concentration visible.)

(c) Pointwise, $(1 + x/n)^n \nearrow \eu^x$ and $x^{1/n} \to
1$ ($x > 0$): the integrand tends to $\eu^{-x}$. Dominator for $n \geq 2$: on $\intoc01$, $x^{-1/n} \leq x^{-1/2}$ and $(1 + x/n)^{-n} \leq 1$: bound $x^{-1/2}$, [integrable](#def-b3-lebesgue-l1); on $\intoo1\infty$, $x^{-1/n} \leq 1$ and $(1 + x/n)^n \geq 1 +
\binom n2\frac{x^2}{n^2} \geq 1 + \frac{x^2}4$: bound $\frac{4}{4 + x^2}$, [integrable](#def-b3-lebesgue-l1). DCT:

$$
\lim_n\int_0^\infty\frac{\dd x}{(1 + x/n)^nx^{1/n}} =
\int_0^\infty\eu^{-x}\dd x = 1 .
$$

## 10.6 Problem: two celebrated integrals

**Problem 10.1.**

Weekend problem — the Gaussian integral and Dirichlet’s integral, by parameters alone

Two integrals rule applied analysis:

$$
G = \int_{-\infty}^{+\infty}\eu^{-x^2}\dd x = \sqrt\pi,
\qquad
D = \int_0^{+\infty}\frac{\sin x}{x}\,\dd x = \frac\pi2
$$

(the second as an improper integral, [Example 10.13](#ex-b3-lebesgue-riemannexamples)). We prove both using only this chapter’s tools.

**Part I — The Gaussian.** For $t \geq 0$ set

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

1. Justify that $A$ and $B$ are $\mathcal C^1$ on $\intoo0{+\infty}$ and compute $A'$ and $B'$ ; show $A'(t) + B'(t) = 0$ . *(In $B'$, substitute $u =  tx$.)*
2. Compute $A(0) + B(0)$ and $\lim_{t\to+\infty}(A + B)(t)$ — justify the limit under the integral in $B$ .
3. Conclude $\int_0^\infty \eu^{-x^2}\dd x =  \frac{\sqrt\pi}2$ , hence $G = \sqrt\pi$ , and deduce $\Gamma(\tfrac12) = \sqrt\pi$ *(substitute $x =  u^2$ in $\Gamma(\frac12)$)* .

**Part II — Dirichlet’s integral.** For $t
\geq 0$ set

$$
F(t) = \int_0^{+\infty}\eu^{-tx}\,\frac{\sin x}{x}\,\dd x .
$$

4. Show that the integral defining $F(t)$ converges for every $t > 0$ as a Lebesgue integral, and for $t = 0$ as an improper integral; show that $D =  \lim_{A\to\infty}\int_0^A\frac{\sin x}x\dd x$ exists *(integrate by parts on $[\pi, A]$)* .
5. Show that $F$ is $\mathcal C^1$ on $\intoo0{+\infty}$ with $$F'(t) = -\int_0^{+\infty}\eu^{-tx}\sin x\,\dd x  = -\frac{1}{1 + t^2}$$ *(domination on $[t_0, \infty)$ for each $t_0 >  0$; the last integral by two integrations by parts or complex exponentials)*.
6. Show $F(t) \to 0$ as $t \to +\infty$ , and deduce $F(t)  = \frac\pi2 - \arctan t$ on $\intoo0{+\infty}$ .
7. The delicate point: $D = \lim_{t\to0^+}F(t)$. Prove it by uniform control of the tail: for $0 \leq t \leq 1$ and $A \geq \pi$, integrate by parts to show $$\Bigl|\int_A^{+\infty}\eu^{-tx}\frac{\sin  x}x\,\dd x\Bigr| \leq \frac{C}{A}$$ with $C$ independent of $t$ *(differentiate $\frac{\eu^{-tx}}x$ and bound $\abs{\cos}$ by $1$; note $t\eu^{-tx} \leq 1/x\cdot(tx\eu^{-tx})$ with $\sup_{u\geq0}u\eu^{-u} < 1$)*; then split $F(t) - D$ into $[0, A]$ (where DCT applies as $t \to 0$) and $[A, \infty)$.
8. Conclude: $D = \frac\pi2$ .

**Part III — Dividends.**

9. Compute $\int_0^{+\infty}\frac{\sin^2x}{x^2}\,\dd x$ *(integrate by parts and reduce to $D$ via $\sin  2x = 2\sin x\cos x$)* .
10. Compute $\int_0^{+\infty}\frac{1 -  \cos x}{x^2}\,\dd x$ , and check the consistency of the two results.
11. For $a > 0$ , compute $\int_0^{+\infty}\frac{\sin(ax)}x\dd x$ and $\int_{-\infty}^{+\infty}\eu^{-ax^2}\dd x$ , and record the scaling rules (they will be the workhorses of [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) ).
12. Explain precisely why $D$ could not have been treated by DCT directly at $t = 0$ (no [integrable](#def-b3-lebesgue-l1) dominator on $[0,1]\times[0,\infty)$ ), and why the tail-splitting of question 7 is the honest substitute — this pattern (“uniform [integrability](#def-b3-lebesgue-l1) of tails”) recurs throughout analysis.

**Part IV — The [Gamma function](#ex-b3-lebesgue-gamma) according to Bohr and Mollerup.** The function $\Gamma$ ([Example 10.16](#ex-b3-lebesgue-gamma)) satisfies $\Gamma(1) = 1$ and $\Gamma(x+1) = x\Gamma(x)$ — but so do infinitely many other functions (multiply by any $1$-periodic wobble). One convexity condition pins $\Gamma$ down uniquely, and its deeper identities then fall out mechanically. A positive function $f$ on an interval is *log-convex* if $\log f$ is convex.

13. Show that log-convex implies convex, that products of log-convex functions and their compositions with affine maps are log-convex, and — via the two-function Hölder inequality $\int\abs{uv} \leq  \bigl(\int\abs u^p\bigr)^{1/p}\bigl(\int\abs  v^q\bigr)^{1/q}$ , proved directly from Young’s inequality — that $\Gamma$ is log-convex on $\intoo0\infty$ .
14. (Slope lemma) Let $g$ be convex on $\intoo0\infty$ with $g(n+1) - g(n) = \log n$ for every integer $n  \geq 1$. For $x \in \intoc01$ and $n \geq 2$, compare the slopes of $g$ over $[n-1, n]$, $[n, n+x]$ and $[n, n+1]$, and deduce $$x\log(n-1) \;\leq\; g(n + x) - g(n) \;\leq\; x\log n .$$
15. (Bohr–Mollerup) Let $f > 0$ satisfy $f(1) = 1$, $f(x+1) = xf(x)$, and $\log f$ convex. Unwinding the recursion into $f(n + x) = x(x+1)\cdots(x + n -  1)\,f(x)$ and $f(n) = (n-1)!$, deduce from question 14 that for $x \in \intoc01$ $$f(x) = \lim_{n\to\infty}  \frac{n!\,n^x}{x(x+1)\cdots(x+n)} :$$ $f$ is unique, hence $f = \Gamma$, and *Gauss’s limit formula* holds (extend to all $x > 0$ by the recursion).
16. Define the *Beta function* $B(x, y) =  \int_0^1t^{x-1}(1-t)^{y-1}\,\dd t$ ( $x, y > 0$ ). Prove convergence, the recursion $B(x+1, y) =  \frac{x}{x+y}\,B(x, y)$ (integrate by parts), and $B(1, y) = \frac1y$ .
17. Show that $x \mapsto B(x, y)$ is log-convex (Hölder again), and apply Bohr–Mollerup to $$f(x) = \frac{B(x, y)\,\Gamma(x + y)}{\Gamma(y)}$$ to conclude *Euler’s formula*: $B(x, y) =  \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ — no double integrals anywhere.
18. Compute $B(\frac12, \frac12)$ directly (substitute $t  = \sin^2\theta$ ) and deduce $\Gamma(\frac12) =  \sqrt\pi$ : the Gaussian integral of Part I, recovered by pure convexity. Compare the two proofs in one sentence each.
19. (Legendre duplication) Show that $$g(x) = \frac{2^{x-1}}{\sqrt\pi}\,  \Gamma\Bigl(\frac x2\Bigr)  \Gamma\Bigl(\frac{x+1}2\Bigr)$$ satisfies the three Bohr–Mollerup hypotheses, and conclude $g = \Gamma$, i.e. $\Gamma(2z) =  \frac{2^{2z-1}}{\sqrt\pi}\,\Gamma(z)\,\Gamma(z +  \tfrac12)$ for all $z > 0$.
20. Deduce the closed form $\Gamma\bigl(n + \tfrac12\bigr)  = \dfrac{(2n)!}{4^n\,n!}\sqrt\pi$, and prove, by the slope lemma applied to $\log\Gamma$ around large integers, the asymptotics $$\frac{\Gamma(n + \frac12)}{\Gamma(n)\,\sqrt n}  \longrightarrow 1 .$$
21. Combine the last two questions into the central binomial asymptotics $$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}},$$ and verify numerically for $n = 10$ ($\binom{20}{10} =  184756$, against $4^{10}/\sqrt{10\pi} \approx  187079$: ratio $\approx 0.988$).
22. (Synthesis) The constant $\sqrt\pi$ has now appeared as the Gaussian integral (Part I), as $B(\frac12,  \frac12)$ (question 18), and inside duplication (question 19); the central binomial estimate anticipates both Stirling ( [Chapter 11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ch-b3-product) ’s weekend problem) and de Moivre–Laplace. Map the connections: which statements are equivalent to which, and what does each technique — differentiation under the integral versus convexity — contribute that the other cannot?

**Part V — Three more dividends.**

23. (Wallis, by Beta) For $p > -1$, substitute $t =  \sin^2\theta$ to show $$W_p = \int_0^{\pi/2}\sin^p\theta\,\dd\theta  = \frac12\,B\Bigl(\frac{p+1}2, \frac12\Bigr),$$ and deduce from the Beta recursion (question 16) that $W_{n+2} = \frac{n+1}{n+2}\,W_n$ for integers $n \geq  0$. Compute $W_{2n}$ and $W_{2n+1}$ in closed form, show $W_{2n+1}/W_{2n} \to 1$ by squeezing, and conclude with *Wallis’s product* $$\frac\pi2 = \lim_{n\to\infty}  \prod_{k=1}^{n}\frac{4k^2}{4k^2 - 1} .$$
24. (The Gaussian meets a frequency) For $b \in \R$ set $$\Phi(b) = \int_{-\infty}^{+\infty}  \eu^{-x^2}\cos(2bx)\,\dd x .$$ Show that $\Phi$ is $\mathcal C^1$ on $\R$, that an integration by parts yields the differential equation $\Phi'(b) = -2b\,\Phi(b)$, and conclude $$\Phi(b) = \sqrt\pi\,\eu^{-b^2} :$$ the Gaussian reproduces itself under this transform — the single identity on which [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) will run.
25. (Frullani’s integral) For $0 < a < b$, show that $$\int_0^{+\infty}  \frac{\eu^{-ax} - \eu^{-bx}}{x}\,\dd x  = \log\frac ba ,$$ by differentiating in the parameter $a$ (justify the domination on every $\intco{a_0}{+\infty}$, $a_0 > 0$, and identify the constant by letting $a \to b$). Where exactly does the integrand need its removable singularity at $x = 0$?

**Solution of Problem 10.1.**

**1.** $A$ is $\mathcal C^1$ by the fundamental theorem of calculus and the chain rule: $A'(t) =
2\eu^{-t^2}\int_0^t\eu^{-x^2}\dd x$. For $B$: $\partial_t\bigl(\frac{\eu^{-t^2(1+x^2)}}{1+x^2}\bigr) =
-2t\,\eu^{-t^2(1+x^2)}$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and bounded on $[t_0, T]
\times \intcc01$ for any $0 < t_0 < T$ (bounded domination on a finite [measure space](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) suffices): $B$ is $\mathcal C^1$ on $\intoo0{+\infty}$ with

$$
B'(t) = -2t\int_0^1 \eu^{-t^2(1+x^2)}\dd x
= -2\eu^{-t^2}\int_0^1 t\,\eu^{-(tx)^2}\dd x
= -2\eu^{-t^2}\int_0^t\eu^{-u^2}\dd u = -A'(t)
$$

(substitution $u = tx$).

**2.** $A(0) = 0$ and $B(0) = \int_0^1\frac{\dd x}{1+x^2}
= \frac\pi4$. $A + B$ has zero derivative on $\intoo0{+\infty}$ and is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $0$ ($B$ by domination $\frac1{1+x^2}$ and [Theorem 10.14](#thm-b3-lebesgue-paramcont)): $A + B \equiv \frac\pi4$. As $t\to+\infty$: $0 \leq B(t) \leq \eu^{-t^2} \to 0$, and $A(t) \to \bigl(\int_0^\infty\eu^{-x^2}\dd x\bigr)^2$ (MCT or just monotone convergence of the inner integral).

**3.** Hence $\bigl(\int_0^\infty\eu^{-x^2}\dd x\bigr)^2
= \frac\pi4$: $\int_0^\infty\eu^{-x^2}\dd x =
\frac{\sqrt\pi}2$, and by evenness $G = \sqrt\pi$. Also $\Gamma(\frac12) = \int_0^\infty x^{-1/2}\eu^{-x}\dd x
\overset{x = u^2}{=} 2\int_0^\infty\eu^{-u^2}\dd u =
\sqrt\pi$.

**4.** For $t > 0$: $\abs{\eu^{-tx}\frac{\sin x}x} \leq
\eu^{-tx}$, [integrable](#def-b3-lebesgue-l1). For $t = 0$, improper convergence: on $[\pi, A]$,

$$
\int_\pi^A\frac{\sin x}x\dd x
= \Bigl[-\frac{\cos x}x\Bigr]_\pi^A -
\int_\pi^A\frac{\cos x}{x^2}\dd x,
$$

both terms convergent as $A \to \infty$; near $0$ the integrand extends [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) by $1$.

**5.** On $[t_0, +\infty)$ ($t_0 > 0$): $\abs{\partial_t\bigl(\eu^{-tx}\tfrac{\sin x}x\bigr)} =
\eu^{-tx}\abs{\sin x} \leq \eu^{-t_0x}$, [integrable](#def-b3-lebesgue-l1): [Theorem 10.15](#thm-b3-lebesgue-paramdiff) applies on every such interval, so on all of $\intoo0{+\infty}$:

$$
F'(t) = -\int_0^\infty\eu^{-tx}\sin x\,\dd x
= -\operatorname{Im}\int_0^\infty\eu^{-(t - \iu)x}\dd x
= -\operatorname{Im}\frac{1}{t - \iu} = -\frac{1}{1 + t^2}.
$$

**6.** $\abs{F(t)} \leq \int_0^\infty\eu^{-tx}\dd x =
\frac1t \to 0$. Integrating $F' = -\frac1{1+t^2}$: $F(t) = C -
\arctan t$, and $t \to \infty$ forces $C = \frac\pi2$: $F(t) =
\frac\pi2 - \arctan t$ on $\intoo0{+\infty}$.

**7.** Integrate by parts on $[A, R]$ with $\sin x =
(-\cos x)'$ and let $R \to \infty$:

$$
\int_A^{\infty}\eu^{-tx}\frac{\sin x}x\dd x
= \frac{\cos A\;\eu^{-tA}}{A}
- \int_A^\infty \cos x\,\Bigl(\frac tx +
\frac1{x^2}\Bigr)\eu^{-tx}\dd x .
$$

Bounding $\abs{\cos} \leq 1$: the first term is $\leq \frac1A$; the integral is at most $\int_A^\infty t\eu^{-tx}\frac{\dd x}x
+ \int_A^\infty\frac{\dd x}{x^2} \leq \frac1A\int_0^\infty
t\eu^{-tx}\dd x + \frac1A = \frac2A$. Total: $\leq \frac 3A$, uniformly for $t \in [0, 1]$ (the case $t = 0$ included). Now

$$
\abs{F(t) - D} \leq
\Bigl|\int_0^A(\eu^{-tx} - 1)\,\frac{\sin x}x\,\dd x\Bigr| +
\frac6A .
$$

On $[0, A]$: $\abs{\eu^{-tx} - 1} \leq tx \leq tA$ and $\abs{\frac{\sin x}x} \leq 1$, so the first term is at most $tA^2$. Choose $A$ with $\frac6A < \varepsilon$, then $t <
\varepsilon/A^2$: $\abs{F(t) - D} < 2\varepsilon$.

**8.** Therefore $D = \lim_{t\to0^+}F(t) =
\lim_{t\to0^+}\bigl(\frac\pi2 - \arctan t\bigr) =
\frac\pi2$.

**9.** By parts ($u = \sin^2x$, $v' = x^{-2}$):

$$
\int_0^\infty\frac{\sin^2x}{x^2}\dd x
= \Bigl[-\frac{\sin^2x}{x}\Bigr]_0^\infty +
\int_0^\infty\frac{2\sin x\cos x}{x}\dd x
= \int_0^\infty\frac{\sin 2x}{x}\dd x = D = \frac\pi2
$$

(substitute $u = 2x$ in the last step; boundary terms vanish: $\sin^2 x/x \to 0$ at both ends).

**10.** By parts ($u = 1 - \cos x$, $v' = x^{-2}$): $\int_0^\infty\frac{1 - \cos x}{x^2}\dd x =
\int_0^\infty\frac{\sin x}x\dd x = \frac\pi2$. Consistency: $1 - \cos x = 2\sin^2\frac x2$, and the substitution $x = 2u$ turns $\int\frac{2\sin^2(x/2)}{x^2}\dd x$ into $\int\frac{\sin^2u}{u^2}\dd u$: the two computations agree.

**11.** $\int_0^\infty\frac{\sin(ax)}x\dd x = \frac\pi2$ for every $a > 0$ (substitute $u = ax$: the integral is scale-invariant); $\int_\R\eu^{-ax^2}\dd x = \sqrt{\pi/a}$ (substitute $u = \sqrt a\,x$). Scaling in the argument leaves the Dirichlet integral fixed and divides the Gaussian by $\sqrt a$.

**12.** A dominator valid for all $t \in [0,1]$ must dominate $\sup_{t\in[0,1]}\abs{\eu^{-tx}\frac{\sin
x}x} = \abs{\frac{\sin x}x}$, which is not [integrable](#def-b3-lebesgue-l1) ([Exercise 10.6](#exo-b3-lebesgue-6)): DCT cannot cross $t = 0$. The substitute of question 7 — tails uniformly small in the parameter, [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) part handled by DCT — is the standard “uniform [integrability](#def-b3-lebesgue-l1)” pattern, and reappears whenever conditional convergence meets limit interchange.

**13.** If $g = \log f$ is convex then $f = \exp\circ g$ is convex (exp is convex increasing: $f(\lambda x + (1-\lambda)
y) \leq \eu^{\lambda g(x) + (1-\lambda)g(y)} \leq \lambda f(x)
+ (1-\lambda)f(y)$, the last step by convexity of exp between the points $g(x), g(y)$). Products and affine substitutions: logarithms turn them into sums and affine substitutions of convex functions. Hölder ($\frac1p + \frac1q = 1$): for $\int\abs u^p = \int\abs v^q = 1$, Young gives $\abs{uv} \leq
\frac{\abs u^p}p + \frac{\abs v^q}q$, integrate: $\int\abs{uv}
\leq 1$; the general case by homogeneity. Then, for $\lambda
\in \intoo01$, apply it with $p = \frac1\lambda$ to the factorization

$$
t^{\lambda x + (1-\lambda)y - 1}\eu^{-t}
= \bigl(t^{x-1}\eu^{-t}\bigr)^{\lambda}
\bigl(t^{y-1}\eu^{-t}\bigr)^{1-\lambda} :
\qquad
\Gamma(\lambda x + (1{-}\lambda)y) \leq
\Gamma(x)^\lambda\,\Gamma(y)^{1-\lambda} .
$$

**14.** For a convex $g$, the slope of a chord increases with its endpoints (three-chord inequality). Comparing the chords over $[n-1, n]$, $[n, n+x]$, $[n, n+1]$:

$$
\log(n-1) = \frac{g(n) - g(n-1)}1 \leq
\frac{g(n+x) - g(n)}x \leq \frac{g(n+1) - g(n)}1 = \log n,
$$

and multiplying by $x > 0$ gives the claim.

**15.** With $f(n) = (n-1)!$ (recursion from $f(1) = 1$) and $f(n + x) = x(x+1)\cdots(x+n-1)\,f(x)$, question 14 reads

$$
(n-1)^x\,(n-1)! \;\leq\; x(x+1)\cdots(x+n-1)\,f(x)
\;\leq\; n^x\,(n-1)! .
$$

The upper bound rewrites as $f(x) \leq
\frac{n!\,n^x}{x(x+1)\cdots(x+n)}\cdot\frac{x+n}n$, and the lower bound at rank $n+1$ as $f(x) \geq
\frac{n!\,n^x}{x(x+1)\cdots(x+n)}$. The correction factor $\frac{x+n}n \to 1$: the sandwich forces

$$
f(x) = \lim_n\frac{n!\,n^x}{x(x+1)\cdots(x+n)}
\qquad (x \in \intoc01),
$$

an expression independent of $f$: uniqueness on $\intoc01$, hence everywhere by the recursion. Since $\Gamma$ satisfies all three hypotheses (question 13), $f = \Gamma$ and Gauss’s formula holds — for all $x > 0$, as both sides obey the same recursion.

**16.** Near $0$, the integrand is $\sim t^{x-1}$, [integrable](#def-b3-lebesgue-l1) iff $x > 0$; near $1$, symmetric with $y$. Integration by parts on $\intcc\varepsilon{1-\varepsilon}$, letting $\varepsilon \to 0$ (boundary terms vanish for $x, y >
0$):

$$
B(x{+}1, y) = \Bigl[-t^x\frac{(1-t)^y}y\Bigr]_0^1 +
\frac xy\int_0^1t^{x-1}(1-t)^y\,\dd t
= \frac xy\bigl(B(x, y) - B(x{+}1, y)\bigr),
$$

using $(1-t)^y = (1-t)^{y-1}(1 - t)$; solving, $B(x+1, y) =
\frac{x}{x+y}B(x, y)$. And $B(1, y) =
\int_0^1(1-t)^{y-1}\dd t = \frac1y$.

**17.** $f(1) = \frac1y\cdot\frac{\Gamma(1+y)}{\Gamma(y)}
= 1$; $f(x+1) = \frac{x}{x+y}B(x,y)\cdot\frac{(x+y)\Gamma(x+y)}
{\Gamma(y)} = x\,f(x)$; and $f$ is log-convex in $x$ as a product of the log-convex $B(\cdot, y)$ (Hölder on the factorization $t^{(\lambda x_1 + (1-\lambda)x_2)-1}(1-t)^{y-1}
= (\cdots)^\lambda(\cdots)^{1-\lambda}$, as in question 13) and $\Gamma(\cdot + y)$ (affine shift). Bohr–Mollerup: $f =
\Gamma$, i.e. $B(x, y) =
\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

**18.** With $t = \sin^2\theta$, $\dd t =
2\sin\theta\cos\theta\,\dd\theta$ and $t^{-1/2}(1-t)^{-1/2} =
\frac1{\sin\theta\cos\theta}$:

$$
B\Bigl(\frac12, \frac12\Bigr) =
\int_0^{\pi/2}2\,\dd\theta = \pi
= \frac{\Gamma(\frac12)^2}{\Gamma(1)}
\quad\Longrightarrow\quad
\Gamma\Bigl(\frac12\Bigr) = \sqrt\pi .
$$

Part I reached the same constant by differentiating a parameter and racing two functions to their limits; here convexity alone rigidified the problem until only one value survived. Analysis by motion versus analysis by shape.

**19.** $g(1) = \frac{2^0}{\sqrt\pi}\Gamma(\frac12)
\Gamma(1) = 1$. Recursion:

$$
g(x+1) = \frac{2^{x}}{\sqrt\pi}\,
\Gamma\Bigl(\frac{x+1}2\Bigr)\Gamma\Bigl(\frac x2 + 1\Bigr)
= \frac{2^{x}}{\sqrt\pi}\cdot\frac x2\,
\Gamma\Bigl(\frac x2\Bigr)\Gamma\Bigl(\frac{x+1}2\Bigr)
= x\,g(x) .
$$

Log-convexity: product of $\eu^{(x-1)\log2}$ (log-affine) and two affine reparametrizations of the log-convex $\Gamma$. Bohr–Mollerup gives $g = \Gamma$; setting $x = 2z$: $\Gamma(2z) = \frac{2^{2z-1}}{\sqrt\pi}\Gamma(z)\Gamma(z +
\frac12)$.

**20.** From $\Gamma(\frac12) = \sqrt\pi$ and the recursion, $\Gamma(n + \frac12) = (n - \frac12)(n -
\frac32)\cdots\frac12\,\sqrt\pi =
\frac{(2n-1)(2n-3)\cdots1}{2^n}\sqrt\pi =
\frac{(2n)!}{4^nn!}\sqrt\pi$ ([complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) the odd product with the evens). Asymptotics: question 14 with $g = \log\Gamma$ and $x = \frac12$ gives $\sqrt{n-1} \leq
\frac{\Gamma(n+\frac12)}{\Gamma(n)} \leq \sqrt n$, so the ratio to $\sqrt n$ is squeezed between $\sqrt{1 - \frac1n}$ and $1$.

**21.** From question 20, $(2n)! =
\frac{4^nn!}{\sqrt\pi}\Gamma(n + \tfrac12)$, so

$$
\binom{2n}n = \frac{(2n)!}{(n!)^2}
= \frac{4^n\,\Gamma(n+\frac12)}{\sqrt\pi\;n!}
= \frac{4^n}{\sqrt\pi}\cdot
\frac{\Gamma(n+\frac12)}{n\,\Gamma(n)}
\sim \frac{4^n}{\sqrt\pi}\cdot\frac{\sqrt n}{n}
= \frac{4^n}{\sqrt{\pi n}} .
$$

Numerically, $4^{10}/\sqrt{10\pi} = 1048576/5.6050 \approx
187078.6$, against $\binom{20}{10} = 184756$: ratio $0.9876$ — the error is $O(1/n)$, visible at $n = 10$.

**22.** Equivalences: $\Gamma(\frac12) = \sqrt\pi
\Leftrightarrow G = \sqrt\pi$ (the substitution $x = u^2$ of question 3) $\Leftrightarrow B(\frac12, \frac12) = \pi$ (Euler’s formula); duplication at $z = n$ *is* the closed form of $\Gamma(n + \frac12)$, which *is* the central binomial estimate up to the slope lemma. The parameter technique (Part I–II) computes limits of moving quantities and is indispensable when a genuine deformation is present (Dirichlet’s integral has no convexity proof); the convexity technique computes nothing but forbids everything — it excels at uniqueness and functional equations (Gauss, Euler, Legendre in three strokes), where differentiation would drown in computation. A [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) analyst carries both.

**23.** With $t = \sin^2\theta$, $\dd t =
2\sin\theta\cos\theta\,\dd\theta =
2\,t^{1/2}(1-t)^{1/2}\,\dd\theta$, so

$$
W_p = \int_0^1 t^{p/2}\,
\frac{\dd t}{2\,t^{1/2}(1-t)^{1/2}}
= \frac12\int_0^1 t^{\frac{p+1}2 - 1}(1-t)^{\frac12 - 1}\dd t
= \frac12\,B\Bigl(\frac{p+1}2, \frac12\Bigr).
$$

The Beta recursion with $x = \frac{n+1}2$, $y = \frac12$ gives

$$
W_{n+2} = \frac12\,
\frac{(n+1)/2}{(n+2)/2}\,B\Bigl(\frac{n+1}2, \frac12\Bigr)
= \frac{n+1}{n+2}\,W_n .
$$

Starting from $W_0 = \frac\pi2$, $W_1 = 1$:

$$
W_{2n} = \frac\pi2\prod_{k=1}^n\frac{2k-1}{2k},
\qquad
W_{2n+1} = \prod_{k=1}^n\frac{2k}{2k+1} .
$$

Since $\sin^{n+1} \leq \sin^n$ on $\intcc0{\pi/2}$, the sequence $(W_n)$ is nonincreasing, so

$$
1 \geq \frac{W_{2n+1}}{W_{2n}} \geq
\frac{W_{2n+1}}{W_{2n-1}} = \frac{2n}{2n+1}
\longrightarrow 1 .
$$

But the closed forms give

$$
\frac{W_{2n+1}}{W_{2n}} = \frac2\pi
\prod_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)}
= \frac2\pi\prod_{k=1}^n\frac{4k^2}{4k^2-1},
$$

and letting $n \to \infty$ yields Wallis’s product. (Via Euler’s formula, $W_p = \frac{\sqrt\pi}2\,
\Gamma(\frac{p+1}2)/\Gamma(\frac p2 + 1)$: Wallis is the Gaussian integral in yet another costume.)

**24.** The $b$-derivative of the integrand is $-2x\,\eu^{-x^2}\sin(2bx)$, dominated by $2\abs
x\,\eu^{-x^2} \in L^1(\R)$ uniformly in $b$: $\Phi$ is $\mathcal C^1$ with

$$
\Phi'(b) = -\int_{-\infty}^{+\infty}
2x\,\eu^{-x^2}\sin(2bx)\,\dd x .
$$

Integrating by parts with $u = \sin(2bx)$, $\dd v =
-2x\,\eu^{-x^2}\dd x$ (so $v = \eu^{-x^2}$), the boundary terms vanish and

$$
\Phi'(b) = -2b\int_{-\infty}^{+\infty}
\eu^{-x^2}\cos(2bx)\,\dd x = -2b\,\Phi(b).
$$

Hence $\bigl(\Phi(b)\,\eu^{b^2}\bigr)' = 0$ and $\Phi(b) =
\Phi(0)\,\eu^{-b^2} = \sqrt\pi\,\eu^{-b^2}$ by Part I. Up to normalization this says the Fourier transform of $\eu^{-x^2}$ is again a Gaussian — the fixed point on which the inversion theory of [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) pivots.

**25.** For $0 < a < b$ and $x > 0$,

$$
0 \leq \frac{\eu^{-ax} - \eu^{-bx}}{x}
= \int_a^b \eu^{-sx}\,\dd s \leq (b - a)\,\eu^{-ax},
$$

so the integral $I(a)$ converges (Lebesgue); the pointwise bound also shows the integrand extends [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) by $b -
a$ at $x = 0$. Fix $b$; on $\intco{a_0}{+\infty}$ the $a$-derivative of the integrand is $-\eu^{-ax}$, dominated by $\eu^{-a_0x} \in L^1(\intoo0{+\infty})$, so $I$ is $\mathcal
C^1$ on $\intoo0b$ with

$$
I'(a) = -\int_0^{+\infty}\eu^{-ax}\dd x = -\frac1a,
\qquad\text{hence}\qquad
I(a) = -\log a + c .
$$

The two-sided bound gives $0 \leq I(a) \leq (b-a)/a \to 0$ as $a \to b^-$, so $c = \log b$ and $I(a) = \log\frac ba$. The removable singularity is needed *at $0$*: each term $\eu^{-ax}/x$ separately has a divergent (logarithmic) integral near $0$, and only the first-order cancellation $\eu^{-ax} - \eu^{-bx} = O(x)$ makes the difference [integrable](#def-b3-lebesgue-l1) there; at infinity each term is harmless on its own.