CommeUnJeu · L2 MP

Parameter-dependent integrals

⌚ ~75 min ▢ 9 blocks ✓ 14 exercises ➣ Prerequisites : Integration on an arbitrary interval, Sequences and series of functions, Numerical and vector series

The chapter Improper integrals on an arbitrary interval built the integration theory of piecewise continuous functions on a general interval $I \subset \mathbb{R}$: convergence, integrability ($L^1(I, \mathbb{K})$), absolute convergence, integration of $O$, $o$, $\sim$ relations. The chapter Sequences and series of functions taught how to exchange limit and integral when the convergence is uniform on a segment. On an arbitrary interval, however, uniform convergence is often unavailable --- the simplest "moving hump" $f_n = n\,\mathbf{1}_{]0, 1/n]}$ converges simply to $0$ on $[0, 1]$ yet $\int_0^1 f_n = 1$ for every $n$. The present chapter delivers two tools that survive on a general $I$: the dominated convergence theorem (CD), where uniformity is replaced by an integrable common dominant, and term-by-term integration (ITTT), the series analogue. From these two tools alone, regularity of a function $g(x) = \int_I f(x, t)\,\mathrm{d}t$ follows: continuity, class $\mathcal{C}^1$, class $\mathcal{C}^k$, and asymptotic studies.
The chapter has four sections. Section~1 states CD and its continuous-parameter extension. Section~2 states ITTT in its positive-functions form and in its signed/complex form with the absolute-integrability assumption $\sum \int_I \lvert f_n \rvert < +\infty$. Section~3 turns to $g(x) = \int_I f(x, t)\,\mathrm{d}t$ and proves continuity from a domination of $\lvert f(x, t) \rvert$. Section~4 proves the Leibniz rule $g'(x) = \int_I \partial f / \partial x (x, t)\,\mathrm{d}t$, extends it to class $\mathcal{C}^k$, and closes with the standard recipe for the asymptotic study of a parameter integral. Hermite's famous late-19th-century anxiety facing the pathological functions of his time --- « je me détourne avec effroi et horreur de cette plaie lamentable des fonctions continues qui n'ont pas de dérivée » (Hermite to Stieltjes, May~1893: I turn away with fright and horror from this lamentable plague of continuous functions that have no derivative) --- sets the stage for the Lebesgue-era tools that this chapter introduces and which resolve, in the prépa setting, the limit-passage tensions Hermite could only suspect.
Standing notation. Throughout, $I$ and $J$ denote intervals of $\mathbb{R}$, with $I$ the integration-variable interval. The parameter is named $x$ throughout the chapter, whether it runs in $\mathbb{N}$ (§1.1), in a real interval $J \subset \mathbb{R}$ (§1.2 and §4), or in $A$ (§3, where $A$ is a part of a finite-dimensional $\mathbb{R}$-normed space). All functions take values in $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$. "p.m." abbreviates piecewise continuous. Following the program's convention, the chapter does not re-state the piecewise-continuity hypothesis in $t$ at every routine Example: it is verified once when the integrand is presented and silently reused. The theorems of Sections~3 and~4 are stated, by default, in their local-domination form, since this is the form used in every CPGE example; the global-domination form is recorded as the simpler limit case.

I Dominated convergence

I.1 The dominated convergence theorem

On a segment $[a, b]$, the chapter Sequences and series of functions allowed the exchange of limit and integral under uniform convergence of $(f_n)$ to $f$. On an arbitrary interval $I$, uniform convergence is too strong: the moving hump $f_n = n\,\mathbf{1}_{]0, 1/n]}$ converges simply to $0$ on $[0, 1]$ but $\norme{f_n - 0}_\infty = n$ for every $n$, so the convergence is far from uniform. The dominated convergence theorem trades uniformity for an integrable common dominant $\varphi$: $\lvert f_n \rvert \le \varphi$ for every $n$, $\varphi$ integrable. The conclusion is exactly what the segment case gave --- $\int_I f_n \to \int_I f$.

Theorem — Dominated convergence theorem

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $f_n \colon I \to \mathbb{K}$. Assume:

each $f_n$ is p.m. on $I$;
$(f_n)$ converges simply on $I$ to a function $f \colon I \to \mathbb{K}$ which is p.m. on $I$;
there exists $\varphi \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall n \in \mathbb{N},\ \forall t \in I,\ \lvert f_n(t) \rvert \le \varphi(t)$.

Then every $f_n$ and $f$ are integrable on $I$ and $$ \int_I f_n(t)\,\mathrm{d}t \xrightarrow[n \to +\infty]{} \int_I f(t)\,\mathrm{d}t. $$

Remark. La démonstration est hors programme. The proof relies on the Lebesgue theory of integration, beyond the reach of this chapter. Two consolations: (i) the theorem itself is among the most powerful in classical analysis --- its hypotheses are remarkably economical (simple convergence + domination), and (ii) the program also marks « L'étude des intégrales semi-convergentes n'est pas un objectif du programme » --- which is consistent here, since CD intrinsically requires absolute integrability through the integrable dominant $\varphi$.

Method — Applying the dominated convergence theorem

The 3-step drill:

check that every $f_n$ is p.m. on $I$;
compute the simple limit $f$ pointwise on $I$ and check that $f$ is p.m.;
exhibit an integrable dominant $\varphi \colon I \to \mathbb{R}_+$ such that $\lvert f_n \rvert \le \varphi$ for every $n$, and verify $\varphi$ is integrable on $I$.

Then $\int_I f_n \to \int_I f$. Watch out: the dominant $\varphi$ must not depend on $n$.

Example

Show that $I_n = \displaystyle\int_1^{+\infty} \frac{1 + t^n}{1 + t^{n+2}}\,\mathrm{d}t \xrightarrow[n \to +\infty]{} 1$.
Set $f_n(t) = (1 + t^n)/(1 + t^{n+2})$ on $I = [1, +\infty[$. Each $f_n$ is continuous on $I$, hence p.m. For $t > 1$, $t^n \to +\infty$ so $f_n(t) \sim t^n / t^{n+2} = 1/t^2$; for $t = 1$, $f_n(1) = 2/2 = 1 = 1/1^2$. Thus $f_n(t) \to f(t) = 1/t^2$ on $[1, +\infty[$, with $f$ continuous on $I$. For the dominant, on $[1, +\infty[$ one has $t^n \le t^{n+2}$, so $1 + t^n \le 1 + t^{n+2} \le 2 t^{n+2}$ (since $1 \le t^{n+2}$), hence $$ \lvert f_n(t) \rvert = \frac{1 + t^n}{1 + t^{n+2}} \le \frac{2 t^{n+2}}{t^{n+2}} = 2 \quad \text{but also} \quad \frac{1 + t^n}{1 + t^{n+2}} \le \frac{2 t^n}{t^{n+2}} = \frac{2}{t^2}. $$ The dominant $\varphi(t) = 2/t^2$ is integrable on $[1, +\infty[$ (Riemann reference, $\alpha = 2 > 1$). By the dominated convergence theorem, $$ I_n = \int_1^{+\infty} f_n(t)\,\mathrm{d}t \xrightarrow[n \to +\infty]{} \int_1^{+\infty} \frac{1}{t^2}\,\mathrm{d}t = 1. $$

Example

Wallis-style: $W_n = \displaystyle\int_0^{\pi/2} \sin^n(t)\,\mathrm{d}t \xrightarrow[n \to +\infty]{} 0$.
Set $f_n(t) = \sin^n(t)$ on $I = [0, \pi/2]$. Each $f_n$ is continuous on $I$. For $t \in [0, \pi/2[$, $0 \le \sin(t) < 1$, hence $f_n(t) \to 0$; for $t = \pi/2$, $\sin(t) = 1$ and $f_n(\pi/2) = 1$ for every $n$. The simple limit is $$ f(t) = \begin{cases} 0 & \text{if } t \in [0, \pi/2[, \\ 1 & \text{if } t = \pi/2, \end{cases} $$ which is piecewise continuous on $[0, \pi/2]$. The constant $\varphi \equiv 1$ dominates every $f_n$ on $[0, \pi/2]$ (since $0 \le \sin t \le 1$), and $\varphi$ is integrable on the segment $[0, \pi/2]$. By CD, $$ W_n \xrightarrow[n \to +\infty]{} \int_0^{\pi/2} f(t)\,\mathrm{d}t = 0 $$ (the value at the single point $t = \pi/2$ contributes $0$ to the integral).

Counter-example 1 (the moving hump). The dominance hypothesis is essential: without it the convergence $\int_I f_n \to \int_I f$ fails (here every $f_n$ and the limit happen to be integrable, but the integrals do not converge to the integral of the limit). Consider $f_n(t) = e^{-t}\,t^n / n!$ on $\mathbb{R}_+$. Each $f_n$ is continuous on $\mathbb{R}_+$ and integrable (Gamma-type decay at $+\infty$). For every fixed $t \ge 0$, by Stirling-free elementary comparisons $t^n / n! \to 0$ as $n \to +\infty$, so $f_n(t) \to 0$ on $\mathbb{R}_+$ --- the simple limit is the zero function. Yet by a direct computation (IPP $n$ times, $\int_0^\infty t^n e^{-t}\,\mathrm{d}t = n!$), $$ \int_0^{+\infty} f_n(t)\,\mathrm{d}t = \frac{1}{n!}\int_0^{+\infty} t^n e^{-t}\,\mathrm{d}t = \frac{n!}{n!} = 1 \quad \text{for every } n, $$ which does not tend to $\int_0^{+\infty} 0\,\mathrm{d}t = 0$. No integrable dominant $\varphi$ exists: any such $\varphi$ would satisfy $\varphi \ge f_n$ pointwise for every $n$, but the $f_n$ pile up a "moving hump" of constant area $1$ that drifts to the right (the peak of $f_n$ is near $t = n$ with value approximately $1/\sqrt{2\pi n}$), so the upper envelope $\sup_n f_n$ is not integrable.

The successive humps drift to the right while keeping a constant area $1$ underneath each: this is the geometric obstruction to dominated convergence. (The curves are vertically rescaled for legibility; the true $f_n$ each enclose unit area.)

Counter-example 2 (the scaled indicator). Define $f_n(t) = n\,\mathbf{1}_{]0, 1/n]}(t)$ on $[0, 1]$ (open on the left so that $f_n(0) = 0$ for every $n$). Each $f_n$ is piecewise continuous on $[0, 1]$. For $t = 0$, $f_n(0) = 0$ for every $n$; for $t \in ]0, 1]$, as soon as $n > 1/t$ one has $t > 1/n$ so $f_n(t) = 0$. Thus $f_n(t) \to 0$ for every $t \in [0, 1]$ --- the simple limit is the zero function. Yet $\int_0^1 f_n(t)\,\mathrm{d}t = n \cdot (1/n) = 1$ for every $n$, which does not tend to $0$. Again no integrable dominant exists: the envelope $\varphi(t) = \sup_n f_n(t)$ equals $\lfloor 1/t \rfloor$, finite at each $t > 0$ but $\sim 1/t$ as $t \to 0^+$, hence not integrable on $]0, 1]$.

Skills to practice

Applying the dominated convergence theorem

I.2 Continuous-parameter extension

The same statement carries over when the discrete index $n \in \mathbb{N}$ is replaced by a continuous parameter $x$ running in an interval $J \subset \mathbb{R}$ and tending to an adherent point $x_0$ of $J$ (possibly $\pm\infty$). The proof is short and lives inside this chapter: it reduces to the sequential case via the sequential characterisation of limits in $\mathbb{R}$ (resp.~in $\overline{\mathbb{R}}$).

Theorem — Dominated convergence -- continuous parameter

Let $J \subset \mathbb{R}$ be an interval and $(f_x)_{x \in J}$ a family of functions $f_x \colon I \to \mathbb{K}$. Let $x_0$ be a point adherent to $J$ in $\overline{\mathbb{R}} = \mathbb{R} \cup \{\pm\infty\}$ (the $\pm\infty$ case is automatically included when $J$ is unbounded). Assume:

for every $x \in J$, $f_x$ is p.m. on $I$;
for every $t \in I$, $f_x(t) \xrightarrow[x \to x_0]{} \ell(t)$ where $\ell \colon I \to \mathbb{K}$ is p.m. on $I$;
there exists $\varphi \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in J,\ \forall t \in I,\ \lvert f_x(t) \rvert \le \varphi(t)$.

Then every $f_x$ and $\ell$ are integrable on $I$ and $$ \int_I f_x(t)\,\mathrm{d}t \xrightarrow[x \to x_0]{} \int_I \ell(t)\,\mathrm{d}t. $$

Proof

By the sequential characterisation of limits in $\overline{\mathbb{R}}$, it suffices to show that for every sequence $(x_n)_{n \in \mathbb{N}}$ of $J$ with $x_n \to x_0$, $\int_I f_{x_n}(t)\,\mathrm{d}t \to \int_I \ell(t)\,\mathrm{d}t$. Fix such a sequence and set $g_n = f_{x_n}$. Then:

each $g_n = f_{x_n}$ is p.m. on $I$ (first hypothesis applied to $x = x_n$);
for every $t \in I$, $g_n(t) = f_{x_n}(t) \to \ell(t)$ (second hypothesis, the limit being independent of the chosen sequence);
$\lvert g_n \rvert = \lvert f_{x_n} \rvert \le \varphi$ on $I$ for every $n$ (third hypothesis applied to $x = x_n$).

The sequential CD (§1.1) applies to $(g_n)$ and yields $\int_I g_n \to \int_I \ell$. Since this holds for every sequence $(x_n) \to x_0$ in $J$, the sequential characterisation of limits gives $\int_I f_x(t)\,\mathrm{d}t \xrightarrow[x \to x_0]{} \int_I \ell(t)\,\mathrm{d}t$.

Example

Show that $\displaystyle\int_0^{+\infty} \frac{e^{-xt}}{1 + t^2}\,\mathrm{d}t \xrightarrow[x \to +\infty]{} 0$.
Set $f_x(t) = e^{-xt}/(1 + t^2)$ on $I = [0, +\infty[$ for $x \in J = [0, +\infty[$. Each $f_x$ is continuous on $I$. We split at $t$: for $t > 0$, $e^{-xt} \to 0$ as $x \to +\infty$, so $f_x(t) \to 0$; for $t = 0$, $f_x(0) = 1$ for every $x$, so the limit at $t = 0$ is $1$. The simple limit is $$ \ell(t) = \begin{cases} 1 & \text{if } t = 0, \\ 0 & \text{if } t > 0, \end{cases} $$ which is piecewise continuous on $[0, +\infty[$. For the dominant, since $e^{-xt} \le 1$ for all $x \ge 0$ and $t \ge 0$, $\lvert f_x(t) \rvert \le 1/(1 + t^2) =: \varphi(t)$, and $\varphi$ is integrable on $[0, +\infty[$ ($\int_0^\infty \mathrm{d}t/(1 + t^2) = \pi/2$). By the continuous CD, $$ \int_0^{+\infty} \frac{e^{-xt}}{1 + t^2}\,\mathrm{d}t \xrightarrow[x \to +\infty]{} \int_0^{+\infty} \ell(t)\,\mathrm{d}t = 0 $$ (the value at the single point $t = 0$ contributes $0$).

Skills to practice

Computing a limit of integrals

II Term-by-term integration

II.1 The positive case

The series analogue of dominated convergence: when can one exchange $\sum$ and $\int_I$? When all $f_n \ge 0$, the answer is the cleanest possible --- the exchange is always legitimate as an equality in $[0, +\infty]$, with both sides simultaneously finite or simultaneously $+\infty$. Following the program's convention, this paragraph does not re-state the piecewise-continuity hypothesis in $t$ for routine examples: « Pour l'application pratique des énoncés de ce paragraphe, on vérifie les hypothèses de convergence simple et de positivité ou de sommabilité, sans expliciter celles relatives à la continuité par morceaux par rapport à $t$. »

Theorem — Term-by-term integration -- positive case

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $f_n \colon I \to \mathbb{R}_+$. Assume:

each $f_n$ is p.m. and integrable on $I$;
$\sum f_n$ converges simply on $I$, with sum $S(t) = \sum_{n=0}^{+\infty} f_n(t)$ p.m. on $I$.

Then the following equality holds in $[0, +\infty]$: $$ \int_I \sum_{n=0}^{+\infty} f_n(t)\,\mathrm{d}t = \sum_{n=0}^{+\infty} \int_I f_n(t)\,\mathrm{d}t. $$ The operational form: the sum $S$ is integrable on $I$ if and only if $\displaystyle\sum_{n=0}^{+\infty} \int_I f_n(t)\,\mathrm{d}t < +\infty$, and in that case the equality above holds in $\mathbb{R}_+$.

Remark. La démonstration est hors programme. As with the dominated convergence theorem, the proof belongs to the Lebesgue theory; the statement above is what the student uses, not what the student proves.

Example

Show that $\displaystyle\int_0^1 \frac{-\ln(t)}{1 - t}\,\mathrm{d}t = \frac{\pi^2}{6}$.
For $t \in ]0, 1[$, the geometric series gives $1/(1 - t) = \sum_{n \ge 0} t^n$, so $$ \frac{-\ln(t)}{1 - t} = -\ln(t)\,\sum_{n=0}^{+\infty} t^n = \sum_{n=0}^{+\infty} f_n(t) \quad \text{with} \quad f_n(t) = -t^n \ln(t). $$ For $t \in ]0, 1[$, $-\ln(t) > 0$ and $t^n \ge 0$, so $f_n \ge 0$ on $]0, 1[$. Each $f_n$ is continuous on $]0, 1[$ and integrable on it: IPP with $u = -\ln(t)$, $\mathrm{d}v = t^n\,\mathrm{d}t$ gives $$ \int_0^1 -t^n \ln(t)\,\mathrm{d}t = \left[ \frac{-t^{n+1} \ln(t)}{n+1} \right]_0^1 + \frac{1}{n+1} \int_0^1 t^n\,\mathrm{d}t = 0 + \frac{1}{(n+1)^2} = \frac{1}{(n+1)^2}. $$ The simple sum $S(t) = -\ln(t)/(1 - t)$ is continuous on $]0, 1[$: it extends continuously at $t = 1$ ($S(t) \to 1$ as $t \to 1^-$), and although $S(t) \to +\infty$ as $t \to 0^+$, it stays integrable near $0$ since $S(t) \sim -\ln(t)$ there, with $-\ln(t)$ integrable on $]0, 1]$. All hypotheses are met. By term-by-term integration (positive case), $$ \int_0^1 \frac{-\ln(t)}{1 - t}\,\mathrm{d}t = \sum_{n=0}^{+\infty} \frac{1}{(n+1)^2} = \sum_{k=1}^{+\infty} \frac{1}{k^2} = \frac{\pi^2}{6}. $$ The last identity is the well-known sum recalled from Series of numbers and vectors. The right-hand side is finite, so the integrand is integrable on $]0, 1[$ (operational form).

Skills to practice

Integrating a positive series term by term

II.2 The signed and complex case

When the $f_n$ change sign (or are complex-valued), simple convergence alone is not enough: there exist series for which $\sum f_n$ converges simply to a function $S$, every $f_n$ is integrable, yet $S$ is not integrable on $I$, or the term-by-term identity fails. The fix is to add the absolute-integrability hypothesis $\sum \int_I \lvert f_n \rvert < +\infty$. This parallels exactly the theory of summable families: « On met en évidence le parallélisme de cet énoncé et du précédent avec ceux issus de la théorie des familles sommables. » (program, p.~21).

Theorem — Term-by-term integration -- signed/complex case

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $f_n \colon I \to \mathbb{K}$. Assume:

each $f_n$ is p.m. and integrable on $I$;
$\sum f_n$ converges simply on $I$, with sum $S(t) = \sum_{n=0}^{+\infty} f_n(t)$ p.m. on $I$;
$\displaystyle\sum_{n=0}^{+\infty} \int_I \lvert f_n(t) \rvert\,\mathrm{d}t < +\infty$ (absolute-integrability).

Then $S$ is integrable on $I$ and $$ \int_I \sum_{n=0}^{+\infty} f_n(t)\,\mathrm{d}t = \sum_{n=0}^{+\infty} \int_I f_n(t)\,\mathrm{d}t. $$

Remark. La démonstration est hors programme. The proof again belongs to the Lebesgue theory and rests on the absolute-integrability assumption combined with the positive-case theorem. The parallel with summable families is structural --- it is the same Fubini-type exchange of sum and integral, formulated in the integration setting.

Method — When term-by-term integration fails: dominated convergence on partial sums

When the absolute-integrability hypothesis $\sum \int_I \lvert f_n \rvert < +\infty$ is in default, the signed-ITTT theorem does not apply directly. The program writes: « On présente des exemples sur lesquels cet énoncé ne s'applique pas, mais dans lesquels l'intégration terme à terme peut être justifiée par le théorème de convergence dominée pour les sommes partielles. » The recipe:

write the partial sum $S_N(t) = \displaystyle\sum_{n=0}^{N} f_n(t)$;
dominate $(S_N)_N$ by a single integrable $\varphi \colon I \to \mathbb{R}_+$ (often via an alternating-series bound or a telescoping bound that exploits sign cancellation);
apply the dominated convergence theorem (§1.1) to $(S_N)$: $\int_I S_N \to \int_I S$;
read $\int_I S_N = \sum_{n=0}^{N} \int_I f_n$ by linearity, then pass to the limit $N \to +\infty$ on both sides to obtain the term-by-term identity.

Example

Show that $\displaystyle\int_0^{+\infty} \frac{\sin(t)}{e^t - 1}\,\mathrm{d}t = \sum_{n=1}^{+\infty} \frac{1}{n^2 + 1}$.
For $t > 0$, $e^t > 1$ so $1/(e^t - 1) = e^{-t}/(1 - e^{-t}) = e^{-t} \sum_{k \ge 0} e^{-kt} = \sum_{n \ge 1} e^{-nt}$ (geometric series, valid since $0 < e^{-t} < 1$ for $t > 0$). Hence $$ \frac{\sin(t)}{e^t - 1} = \sum_{n=1}^{+\infty} f_n(t) \quad \text{with} \quad f_n(t) = e^{-nt} \sin(t). $$ Each $f_n$ is continuous on $]0, +\infty[$ and integrable: $\lvert f_n(t) \rvert \le e^{-nt}$, integrable on $]0, +\infty[$ with integral $1/n$.
Absolute-integrability check. Using the global bound $\lvert \sin t \rvert \le t$ valid for $t \ge 0$, $\lvert f_n(t) \rvert = e^{-nt} \lvert \sin t \rvert \le t\,e^{-nt}$, and $\int_0^\infty t\,e^{-nt}\,\mathrm{d}t = 1/n^2$ (IPP). So $\sum_{n \ge 1} \int_0^\infty \lvert f_n \rvert \le \sum_{n \ge 1} 1/n^2 < +\infty$ --- the signed-ITTT hypothesis is met.
Computation of $\int_0^\infty f_n$. Either by IPP twice on $\int_0^\infty e^{-nt} \sin t\,\mathrm{d}t$ or by writing $\sin t = \mathrm{Im}(e^{it})$ and computing $\int_0^\infty e^{-nt} e^{it}\,\mathrm{d}t = 1/(n - i) = (n + i)/(n^2 + 1)$, the imaginary part gives $\int_0^\infty e^{-nt} \sin t\,\mathrm{d}t = 1/(n^2 + 1)$.
By term-by-term integration (signed case), $$ \int_0^{+\infty} \frac{\sin(t)}{e^t - 1}\,\mathrm{d}t = \sum_{n=1}^{+\infty} \frac{1}{n^2 + 1}. $$

Example

Counter-example to absolute ITTT, rescued by CD on partial sums: compute $\displaystyle\int_0^1 \frac{\mathrm{d}t}{\sqrt{t}\,(1 + t)} = \frac{\pi}{2}$ and the byproduct $\displaystyle\sum_{n=0}^{+\infty} \frac{(-1)^n}{n + 1/2} = \frac{\pi}{2}$.
On $]0, 1[$, $1/(1 + t) = \sum_{n \ge 0} (-1)^n t^n$ (geometric series), so $$ \frac{1}{\sqrt{t}\,(1 + t)} = \sum_{n=0}^{+\infty} f_n(t) \quad \text{with} \quad f_n(t) = (-1)^n t^{n - 1/2}. $$ Each $f_n$ is continuous on $]0, 1[$ and integrable: $\int_0^1 \lvert f_n \rvert\,\mathrm{d}t = \int_0^1 t^{n - 1/2}\,\mathrm{d}t = 1/(n + 1/2)$.
Absolute-integrability fails: $\displaystyle\sum_{n=0}^{+\infty} \int_0^1 \lvert f_n \rvert = \sum_{n=0}^{+\infty} \frac{1}{n + 1/2}$ diverges (harmonic-type series, since $1/(n + 1/2) \sim 1/n$). So the signed-ITTT theorem does not apply directly.
Rescue by CD on partial sums. Write $S_N(t) = \sum_{n=0}^{N} (-1)^n t^{n - 1/2} = t^{-1/2}\sum_{n=0}^{N} (-t)^n$. For $t \in ]0, 1[$, the general term $a_n = t^n$ of the inner alternating sum is positive decreasing to $0$, so by the CSSA partial-sum bracketing (recalled from Series of numbers and vectors), $0 < 1 - t \le \sum_{n=0}^{N} (-t)^n \le 1$ for every $N$, hence $$ \lvert S_N(t) \rvert \le t^{-1/2} \cdot 1 = t^{-1/2}. $$ The dominant $\varphi(t) = t^{-1/2}$ is integrable on $]0, 1[$ (Riemann-type, $\alpha = 1/2 < 1$).
By CD applied to $(S_N)$, $\int_0^1 S_N(t)\,\mathrm{d}t \to \int_0^1 t^{-1/2}/(1 + t)\,\mathrm{d}t$ as $N \to +\infty$. By linearity $\int_0^1 S_N = \sum_{n=0}^N (-1)^n/(n + 1/2)$, so passing to the limit $$ \int_0^1 \frac{\mathrm{d}t}{\sqrt{t}\,(1 + t)} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n + 1/2}. \tag{$\star$} $$ Closed-form evaluation by elementary substitution. Set $u = \sqrt{t}$, so $t = u^2$ and $\mathrm{d}t = 2u\,\mathrm{d}u$; the bounds $t \in ]0, 1[$ become $u \in ]0, 1[$. Then $$ \int_0^1 \frac{\mathrm{d}t}{\sqrt{t}\,(1 + t)} = \int_0^1 \frac{2u\,\mathrm{d}u}{u\,(1 + u^2)} = 2\int_0^1 \frac{\mathrm{d}u}{1 + u^2} = 2 \arctan(1) - 2\arctan(0) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}. $$ Combining with~$(\star)$, $$ \sum_{n=0}^{+\infty} \frac{(-1)^n}{n + 1/2} = \frac{\pi}{2}. $$ This is the byproduct: the term-by-term identity plus an elementary closed-form integration delivers a non-trivial alternating-series identity without any boundary-of-disc series-power machinery.

Skills to practice

Integrating a signed series term by term

III Continuity of a parameter integral

III.1 Global and local domination

The goal of the section is to study $g(x) = \int_I f(x, t)\,\mathrm{d}t$ where $f \colon A \times I \to \mathbb{K}$ and $A$ is a part of a finite-dimensional $\mathbb{R}$-normed space. Two questions: (i) for which $x \in A$ is $g(x)$ defined? (ii) is $g$ continuous on $A$? Following the program's convention, hypotheses of piecewise-continuity in $t$ are stated once and not re-checked at every Example: « Pour l'application pratique des énoncés de ce paragraphe, on vérifie les hypothèses de régularité par rapport à $x$ et de domination, sans expliciter celles relatives à la continuité par morceaux par rapport à $t$. »

Theorem — Continuity under the integral sign -- global domination

Let $A$ be a part of a finite-dimensional $\mathbb{R}$-normed space, $I$ an interval of $\mathbb{R}$, and $f \colon A \times I \to \mathbb{K}$. Assume:

for every $t \in I$, $x \mapsto f(x, t)$ is continuous on $A$;
for every $x \in A$, $t \mapsto f(x, t)$ is p.m. on $I$;
there exists $\varphi \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in A,\ \lvert f(x, \cdot) \rvert \le \varphi$ on $I$.

Then $g \colon x \mapsto \displaystyle\int_I f(x, t)\,\mathrm{d}t$ is defined and continuous on $A$.

Proof

Existence of $g(x)$. For every $x \in A$, $\lvert f(x, \cdot) \rvert \le \varphi$ with $\varphi$ integrable on $I$, hence $t \mapsto f(x, t)$ is integrable on $I$ and $g(x) = \int_I f(x, t)\,\mathrm{d}t$ is defined.
Continuity at $x_0 \in A$. By the sequential characterisation of continuity in a finite-dim normed space, it suffices to show that for every sequence $(x_n)$ of $A$ with $x_n \to x_0$, $g(x_n) \to g(x_0)$. Fix such a sequence and set $g_n(t) = f(x_n, t)$. Then:

each $g_n$ is p.m. on $I$ (second hypothesis with $x = x_n$);
for every $t \in I$, $g_n(t) = f(x_n, t) \to f(x_0, t)$ by continuity of $x \mapsto f(x, t)$ at $x_0$ (first hypothesis); the limit $f(x_0, \cdot)$ is p.m. on $I$ (second hypothesis with $x = x_0$);
$\lvert g_n \rvert = \lvert f(x_n, \cdot) \rvert \le \varphi$ on $I$ (third hypothesis with $x = x_n$).

The sequential CD (§1.1) applies and yields $g(x_n) = \int_I g_n \to \int_I f(x_0, \cdot) = g(x_0)$.

Theorem — Continuity under the integral sign -- local domination

Same setting as above, with the global domination weakened to: on every compact part $K \subset A$, there exists $\varphi_K \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in K,\ \lvert f(x, \cdot) \rvert \le \varphi_K$ on $I$. Then $g \colon x \mapsto \int_I f(x, t)\,\mathrm{d}t$ is defined and continuous on $A$.
Remark (program, p.~21). « En pratique, on vérifie l'hypothèse de domination sur tout segment de $A$, ou sur d'autres intervalles adaptés à la situation. » When $A$ is an interval of $\mathbb{R}$, it suffices to check the domination on every segment $[a, b] \subset A$ --- every compact part of $A$ is contained in such a segment, so segments are the convenient test family.

Proof

Continuity is a local property, so it suffices to check continuity on every compact part $K \subset A$. Fix $K$; the local-domination hypothesis gives an integrable $\varphi_K$ on $I$ such that $\lvert f(x, \cdot) \rvert \le \varphi_K$ for every $x \in K$. The triple of hypotheses needed by the global-domination theorem, applied with $A$ replaced by $K$ and $\varphi$ by $\varphi_K$, is satisfied. Hence $g$ restricted to $K$ is continuous on $K$. Since this holds for every compact part $K \subset A$, $g$ is continuous on $A$.

Method — Local-domination drill

The standard 3-step pattern for applying the local-domination theorem:

fix a compact part $K \subset A$ --- when $A$ is an interval of $\mathbb{R}$ (the typical CPGE case), take $K = [a, b] \subset A$; in general, take any compact $K \subset A$;
for $x \in K$, dominate $\lvert f(x, t) \rvert$ by some $\varphi_K(t)$ not depending on $x$ (typically the worst case over $K$, e.g. $\varphi_K(t) = \sup_{x \in K} \lvert f(x, t) \rvert$ when this supremum is itself integrable, or a clever majorant derived from monotonicity in $x$);
check that $\varphi_K$ is integrable on $I$.

Example

Starter: show that $g(x) = \displaystyle\int_0^{+\infty} e^{-t} \sin(xt)\,\mathrm{d}t$ is defined and continuous on $\mathbb{R}$.
Set $f(x, t) = e^{-t} \sin(xt)$ on $\mathbb{R} \times [0, +\infty[$. For every $t \ge 0$, $x \mapsto e^{-t} \sin(xt)$ is continuous on $\mathbb{R}$ (composition of continuous maps). For every $x \in \mathbb{R}$, $t \mapsto e^{-t} \sin(xt)$ is continuous on $[0, +\infty[$, hence p.m. Global domination: for all $x \in \mathbb{R}$ and $t \ge 0$, $\lvert e^{-t} \sin(xt) \rvert \le e^{-t}$, and $\varphi(t) = e^{-t}$ is integrable on $[0, +\infty[$ with $\int_0^\infty e^{-t}\,\mathrm{d}t = 1$. By the global-domination Theorem, $g$ is defined and continuous on $\mathbb{R}$.

Example

The Gamma function of Euler -- definition and continuity. Define $\Gamma(x) = \displaystyle\int_0^{+\infty} e^{-t}\,t^{x - 1}\,\mathrm{d}t$. Show that $\Gamma$ is defined and continuous on $]0, +\infty[$.
Set $f(x, t) = e^{-t} t^{x - 1}$ on $]0, +\infty[ \times ]0, +\infty[$.
Domain of definition. Fix $x > 0$. The function $t \mapsto e^{-t} t^{x - 1}$ is continuous on $]0, +\infty[$. Problem at $0$: $\int_0^1 t^{x - 1}\,\mathrm{d}t$ is a reference Riemann integral, convergent iff $x - 1 > -1$, i.e. iff $x > 0$ --- satisfied. Problem at $+\infty$: by exponential growth dominating polynomial, $e^{-t} t^{x - 1} = o(1/t^2)$ as $t \to +\infty$, so the integral converges at $+\infty$. Hence $\Gamma(x)$ is well-defined for every $x > 0$.
Continuity by local domination. For $x \mapsto f(x, t) = e^{-t} t^{x - 1}$ to be continuous in $x$ for each $t > 0$: $t^{x - 1} = e^{(x - 1)\ln t}$ is continuous in $x$, so yes. For $t \mapsto f(x, t)$ piecewise continuous in $t$ for each $x > 0$: yes (it is continuous on $]0, +\infty[$). For the domination: fix a segment $[a, b] \subset ]0, +\infty[$ with $0 < a \le b$. For $x \in [a, b]$ and $t > 0$, $$ \lvert e^{-t} t^{x - 1} \rvert = e^{-t} t^{x - 1} \le \varphi_{[a, b]}(t) \quad \text{with} \quad \varphi_{[a, b]}(t) = \begin{cases} e^{-t} t^{a - 1} & \text{if } 0 < t \le 1, \\ e^{-t} t^{b - 1} & \text{if } t \ge 1, \end{cases} $$ because for $0 < t \le 1$, $t^{x - 1} \le t^{a - 1}$ (since $x \ge a$ and $t \le 1$ makes $t \mapsto t^c$ decreasing in $c$), and for $t \ge 1$, $t^{x - 1} \le t^{b - 1}$ (since $x \le b$ and $t \ge 1$ makes $t \mapsto t^c$ increasing in $c$). The dominant $\varphi_{[a, b]}$ is integrable on $]0, +\infty[$ by the domain-of-definition argument applied at $x = a$ and $x = b$. The local-domination Theorem gives $\Gamma$ continuous on every segment $[a, b] \subset ]0, +\infty[$, hence continuous on $]0, +\infty[$.

Skills to practice

Proving continuity of a parameter integral

III.2 The segment case

When the integration interval is a segment $[a, b]$, continuity of $f$ on $A \times [a, b]$ alone suffices --- no domination hypothesis required. The proof rests on the boundedness of continuous functions on compact sets.

Theorem — Continuity under the integral sign -- segment case

Let $A$ be a part of a finite-dimensional $\mathbb{R}$-normed space and $[a, b]$ a segment of $\mathbb{R}$. Let $f \colon A \times [a, b] \to \mathbb{K}$ be continuous on $A \times [a, b]$. Then $g \colon x \mapsto \int_a^b f(x, t)\,\mathrm{d}t$ is continuous on $A$.
The result is directly usable as it stands. Its proof is short enough --- one application of "continuous on a compact $\Rightarrow$ bounded" --- that it is often re-derived inline rather than invoked by name, the bounding constant depending on the compact part of $A$ under consideration.

Proof

The conclusion is local in $A$, so it suffices to check continuity at every $x_0 \in A$. We use the sequential characterisation. Fix $x_0 \in A$ and let $(x_n)$ be any sequence of $A$ with $x_n \to x_0$. The set $K = \{x_0\} \cup \{x_n : n \in \mathbb{N}\}$ is compact in the ambient normed space (a convergent sequence together with its limit is compact). The product $K \times [a, b]$ is then compact, and $f$ is continuous on it, hence bounded by some constant $M_K \ge 0$: $\lvert f(x, t) \rvert \le M_K$ for all $(x, t) \in K \times [a, b]$.
The sequence $g_n(t) = f(x_n, t)$ satisfies: $g_n$ is continuous on $[a, b]$ (hence p.m.); for every $t \in [a, b]$, $g_n(t) = f(x_n, t) \to f(x_0, t)$ by continuity of $f$; the constant $\varphi \equiv M_K$ dominates every $\lvert g_n \rvert$ on $[a, b]$ and is integrable on the segment. By sequential CD (§1.1), $g(x_n) = \int_a^b g_n \to \int_a^b f(x_0, \cdot) = g(x_0)$. Since this holds for every sequence $(x_n) \to x_0$, $g$ is continuous at $x_0$ by the sequential characterisation. Since $x_0$ was arbitrary, $g$ is continuous on $A$.

Example

Show that $g(x) = \displaystyle\int_0^1 \frac{\cos(xt)}{1 + t^2}\,\mathrm{d}t$ is continuous on $\mathbb{R}$.
The integration interval $[0, 1]$ is a segment, and $f(x, t) = \cos(xt)/(1 + t^2)$ is continuous on $\mathbb{R} \times [0, 1]$ as a composition of continuous maps. By the segment-case Theorem, $g$ is continuous on $\mathbb{R}$.

Skills to practice

Using the segment case

IV Class $\mathcal{C}^1$ and class $\mathcal{C}^k$

IV.1 Class $\mathcal{C}^1$ -- the Leibniz theorem

Can one swap $\mathrm{d}/\mathrm{d}x$ and $\int_I$? On a segment with uniform convergence of $\partial f / \partial x$, the chapter Sequences and series of functions delivered "yes". On an arbitrary interval $I$, the natural replacement is again a domination --- this time of $\partial f / \partial x$ by an integrable $\varphi$. The proof is short and instructive: it reduces to the continuous CD theorem (§1.2) applied to the difference quotient $\tau(x, t) = (f(x, t) - f(x_0, t))/(x - x_0)$, with the domination of $\tau$ obtained from the inequality of finite increments. From this point on, the parameter $x$ runs in an interval $A \subset \mathbb{R}$ (the program's restriction for the $\mathcal{C}^1$ statement).

Theorem — Leibniz -- class $\mathcal{C}^1$$\virgule$ global domination

Let $A$ and $I$ be two intervals of $\mathbb{R}$ and $f \colon A \times I \to \mathbb{K}$. Assume:

for every $t \in I$, $x \mapsto f(x, t)$ is of class $\mathcal{C}^1$ on $A$;
for every $x \in A$, $t \mapsto f(x, t)$ is integrable on $I$;
for every $x \in A$, $t \mapsto \partial f / \partial x (x, t)$ is p.m. on $I$;
there exists $\varphi \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in A,\ \lvert \partial f / \partial x (x, \cdot) \rvert \le \varphi$ on $I$.

Then $g \colon x \mapsto \int_I f(x, t)\,\mathrm{d}t$ is of class $\mathcal{C}^1$ on $A$ and $$ \forall x \in A,\quad g'(x) = \int_I \frac{\partial f}{\partial x}(x, t)\,\mathrm{d}t. $$

Proof

Fix $x_0 \in A$ and consider the difference quotient $\tau(x, t) = (f(x, t) - f(x_0, t))/(x - x_0)$ for $x \in A \setminus \{x_0\}$ and $t \in I$.
(a) For each $x \in A \setminus \{x_0\}$, $\tau(x, \cdot) = (f(x, \cdot) - f(x_0, \cdot))/(x - x_0)$ is p.m. on $I$ as a difference of integrable, hence p.m., functions, divided by a nonzero constant.
(b) For each $t \in I$, $\tau(x, t) \xrightarrow[x \to x_0]{} \partial f / \partial x (x_0, t)$ by definition of the partial derivative in $x$ at $x_0$ (first hypothesis); the limit $t \mapsto \partial f / \partial x (x_0, t)$ is p.m. on $I$ (third hypothesis with $x = x_0$).
(c) Domination of $\tau$. For each $t \in I$, $x \mapsto f(x, t)$ is $\mathcal{C}^1$ on $A$ (first hypothesis), and on the segment $[x_0, x]$ (or $[x, x_0]$) its derivative in $x$ is dominated by $\varphi(t)$ (fourth hypothesis). The inequality of finite increments applied to $x \mapsto f(x, t)$ on this segment gives $$ \lvert f(x, t) - f(x_0, t) \rvert \le \lvert x - x_0 \rvert \cdot \sup_{y \in [x_0, x] \cup [x, x_0]} \lvert \partial f / \partial x (y, t) \rvert \le \lvert x - x_0 \rvert \cdot \varphi(t), $$ hence $\lvert \tau(x, t) \rvert \le \varphi(t)$ for every $x \in A \setminus \{x_0\}$ and every $t \in I$. The dominant $\varphi$ is integrable on $I$.
We conclude through the sequential characterisation of the limit at $x_0$ (the domain $A \setminus \{x_0\}$ is not an interval when $x_0$ is interior, so the continuous-parameter statement of §1.2 does not apply verbatim; the sequential theorem of §1.1 does). Let $(x_p)_{p \in \mathbb{N}}$ be any sequence of $A \setminus \{x_0\}$ with $x_p \to x_0$. By (a)--(c), the sequence $\big(\tau(x_p, \cdot)\big)_p$ satisfies the hypotheses of the sequential CD theorem (§1.1): each is p.m. on $I$, the simple limit is $t \mapsto \partial f / \partial x (x_0, t)$ p.m. on $I$, and the common dominant $\varphi$ is integrable. Hence $\int_I \tau(x_p, t)\,\mathrm{d}t \to \int_I \partial f / \partial x (x_0, t)\,\mathrm{d}t$. As this holds for every such sequence, $$ \frac{g(x) - g(x_0)}{x - x_0} = \int_I \tau(x, t)\,\mathrm{d}t \xrightarrow[x \to x_0]{} \int_I \frac{\partial f}{\partial x}(x_0, t)\,\mathrm{d}t. $$ The left-hand side has a finite limit, so $g$ is differentiable at $x_0$ with $g'(x_0) = \int_I \partial f / \partial x (x_0, t)\,\mathrm{d}t$. Since this holds for every $x_0 \in A$, $g$ is differentiable on $A$ with the announced formula. Continuity of $g'$ on $A$ follows from §3.1 (global-domination Theorem applied to $\partial f / \partial x$, whose hypotheses are exactly those of the present Theorem). So $g$ is of class $\mathcal{C}^1$ on $A$.

Theorem — Leibniz -- class $\mathcal{C}^1$$\virgule$ local domination

Same setting as above, with the global domination of $\partial f / \partial x$ weakened to: on every segment $[a, b] \subset A$, there exists $\varphi_{[a, b]} \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in [a, b],\ \lvert \partial f / \partial x (x, \cdot) \rvert \le \varphi_{[a, b]}$ on $I$. Then $g$ is of class $\mathcal{C}^1$ on $A$ with $g'(x) = \int_I \partial f / \partial x (x, t)\,\mathrm{d}t$.
Proof. The conclusions $g \in \mathcal{C}^1(A)$ and $g' = \int_I \partial f / \partial x$ are local, so it suffices to apply the global-domination Theorem on every segment $[a, b] \subset A$ with $\varphi$ replaced by $\varphi_{[a, b]}$.

Method — Applying the Leibniz theorem

The 4-step pattern, with domination already in segment-by-segment form (the form used in every CPGE example):

check that $x \mapsto f(x, t)$ is of class $\mathcal{C}^1$ on $A$ for every $t \in I$;
check that $t \mapsto f(x, t)$ is integrable on $I$ for every $x \in A$;
check that $t \mapsto \partial f / \partial x (x, t)$ is p.m. on $I$ for every $x \in A$;
build the segment-wise domination: fix $[a, b] \subset A$, find $\varphi_{[a, b]}$ integrable on $I$ such that $\forall x \in [a, b],\ \lvert \partial f / \partial x (x, \cdot) \rvert \le \varphi_{[a, b]}$.

Conclude: $g \in \mathcal{C}^1(A)$ with $g'(x) = \int_I \partial f / \partial x (x, t)\,\mathrm{d}t$. Often, this expression of $g'$ leads to a differential equation satisfied by $g$, which then determines $g$ up to a constant.

Example

Continuing the starter example. Show that $g(x) = \displaystyle\int_0^{+\infty} e^{-t} \sin(xt)\,\mathrm{d}t$ is of class $\mathcal{C}^1$ on $\mathbb{R}$ and compute $g'$.
Set $f(x, t) = e^{-t} \sin(xt)$. For every $t \ge 0$, $x \mapsto e^{-t} \sin(xt)$ is $\mathcal{C}^\infty$ on $\mathbb{R}$, in particular $\mathcal{C}^1$. For every $x \in \mathbb{R}$, $t \mapsto e^{-t} \sin(xt)$ is integrable on $[0, +\infty[$ (already verified). $\partial f / \partial x (x, t) = e^{-t} \cdot t \cos(xt)$ is continuous in $t$. Global domination of $\partial f / \partial x$: $\lvert \partial f / \partial x (x, t) \rvert = t\,e^{-t} \lvert \cos(xt) \rvert \le t\,e^{-t} =: \varphi(t)$, and $\varphi$ is integrable on $[0, +\infty[$ (IPP gives $\int_0^\infty t\,e^{-t}\,\mathrm{d}t = 1$). All hypotheses are met. By Leibniz with global domination, $g$ is $\mathcal{C}^1$ on $\mathbb{R}$ with $$ g'(x) = \int_0^{+\infty} t\,e^{-t} \cos(xt)\,\mathrm{d}t. $$

Example

Gamma function -- class $\mathcal{C}^1$. Show that $\Gamma$ is of class $\mathcal{C}^1$ on $]0, +\infty[$ with $\Gamma'(x) = \int_0^{+\infty} \ln(t)\,e^{-t} t^{x - 1}\,\mathrm{d}t$.
Set $f(x, t) = e^{-t} t^{x - 1}$. For every $t > 0$, $x \mapsto e^{-t} t^{x - 1} = e^{-t}\,e^{(x - 1)\ln t}$ is $\mathcal{C}^\infty$ on $\mathbb{R}$ (composition with the exponential), in particular $\mathcal{C}^1$. Its derivative in $x$ is $\partial f / \partial x (x, t) = \ln(t)\,e^{-t} t^{x - 1}$. For every $x > 0$, integrability of $t \mapsto e^{-t} t^{x - 1}$ on $]0, +\infty[$ was established in §3.1. $t \mapsto \ln(t)\,e^{-t} t^{x - 1}$ is continuous on $]0, +\infty[$, hence p.m.
Local domination of $\partial f / \partial x$: fix $[a, b] \subset ]0, +\infty[$. For $x \in [a, b]$ and $t > 0$, $$ \lvert \ln(t)\,e^{-t} t^{x - 1} \rvert = e^{-t}\,t^{x - 1}\,\lvert \ln t \rvert \le \varphi_{[a, b]}(t) \quad \text{with} \quad \varphi_{[a, b]}(t) = \begin{cases} e^{-t}\,t^{a - 1}\,\lvert \ln t \rvert & \text{if } 0 < t \le 1, \\ e^{-t}\,t^{b - 1}\,\lvert \ln t \rvert & \text{if } t \ge 1. \end{cases} $$ The dominant $\varphi_{[a, b]}$ is integrable on $]0, +\infty[$: at $0$, $\lvert \ln t \rvert\,t^{a - 1} = o(t^{a - 1 - \delta})$ for any $\delta \in ]0, a[$, and $\int_0^1 t^{a - 1 - \delta}\,\mathrm{d}t$ converges (Riemann-Bertrand comparison), giving convergence near $0$; at $+\infty$, $\lvert \ln t \rvert\,e^{-t} t^{b - 1} = o(1/t^2)$ by exponential decay. By Leibniz with local domination, $\Gamma$ is $\mathcal{C}^1$ on $]0, +\infty[$ with the announced derivative.

Skills to practice

Applying the Leibniz rule

IV.2 Class $\mathcal{C}^k$

By induction on $k$, the $\mathcal{C}^1$ theorem gives a $\mathcal{C}^k$ analogue. The program's hypothesis is parsimonious: a domination is required only of the top derivative $\partial^k f / \partial x^k$, while the intermediate derivatives $\partial^j f / \partial x^j$ for $0 \le j \le k - 1$ are only required to be integrable. The proof builds dominants for the intermediate derivatives from the top one via iterated inequality-of-finite-increments cascade.

Theorem — Leibniz -- class $\mathcal{C}^k$$\virgule$ local domination

Let $A$ and $I$ be two intervals of $\mathbb{R}$, $k \ge 1$ an integer, and $f \colon A \times I \to \mathbb{K}$. Assume:

for every $t \in I$, $x \mapsto f(x, t)$ is of class $\mathcal{C}^k$ on $A$;
for every $j \in \llbracket 0, k - 1 \rrbracket$ and every $x \in A$, $t \mapsto \partial^j f / \partial x^j (x, t)$ is integrable on $I$;
for every $x \in A$, $t \mapsto \partial^k f / \partial x^k (x, t)$ is p.m. on $I$;
on every segment $[a, b] \subset A$, there exists $\varphi_{[a, b]} \colon I \to \mathbb{R}_+$ integrable on $I$ such that $\forall x \in [a, b],\ \lvert \partial^k f / \partial x^k (x, \cdot) \rvert \le \varphi_{[a, b]}$ on $I$.

Then $g \colon x \mapsto \int_I f(x, t)\,\mathrm{d}t$ is of class $\mathcal{C}^k$ on $A$ and $$ \forall x \in A,\quad g^{(k)}(x) = \int_I \frac{\partial^k f}{\partial x^k}(x, t)\,\mathrm{d}t. $$

Proof

By induction on $k \ge 1$. The base case $k = 1$ is the $\mathcal{C}^1$ Leibniz theorem (local form) of §4.1. Suppose the result holds at order $k - 1$ for some $k \ge 2$, and let $f$ satisfy the order-$k$ hypotheses.
Step 1 -- intermediate dominants. The conclusion is local in $A$, so fix a segment $[a, b] \subset A$ and pick $x_0 \in [a, b]$. The top-derivative domination $\lvert \partial^k f / \partial x^k (x, \cdot) \rvert \le \varphi_{[a, b]}$ for $x \in [a, b]$, combined with the integrability hypotheses on $\partial^j f / \partial x^j (x_0, \cdot)$ for $0 \le j \le k - 1$, lets us construct integrable dominants $\psi_j$ on $I$ for each lower-order derivative as follows. One application of the inequality of finite increments to $x \mapsto \partial^j f / \partial x^j (x, t)$ on $[\min(x_0, x), \max(x_0, x)] \subset [a, b]$ gives $$ \lvert \partial^j f / \partial x^j (x, t) - \partial^j f / \partial x^j (x_0, t) \rvert \le (b - a)\,\sup_{y \in [a, b]} \lvert \partial^{j + 1} f / \partial x^{j + 1} (y, t) \rvert. $$ Iterating this estimate telescopes upward: for every $j \in \llbracket 0, k - 1 \rrbracket$, $$ \sup_{y \in [a, b]} \lvert \partial^j f / \partial x^j (y, t) \rvert \le \sum_{m = j}^{k - 1} (b - a)^{m - j}\,\lvert \partial^m f / \partial x^m (x_0, t) \rvert + (b - a)^{k - j}\,\varphi_{[a, b]}(t). $$ Each term on the right-hand side is integrable in $t$ over $I$ (the second hypothesis gives integrability of $t \mapsto \partial^m f / \partial x^m (x_0, t)$ for $0 \le m \le k - 1$; the fourth gives integrability of $\varphi_{[a, b]}$), so the sum defines an integrable dominant $\psi_j \colon I \to \mathbb{R}_+$ on $[a, b] \times I$ for every $j$.
Step 2 -- applying the inductive hypothesis at order $k - 1$. Define $h \colon A \times I \to \mathbb{K}$ by $h(x, t) = f(x, t)$ --- same function, but inspected with the order-$(k - 1)$ hypotheses. The hypotheses at order $k - 1$ for $h$ read:

$x \mapsto h(x, t)$ is $\mathcal{C}^{k - 1}$ on $A$ (clear from $f$ being $\mathcal{C}^k$);
$t \mapsto \partial^j h / \partial x^j (x, t) = \partial^j f / \partial x^j (x, t)$ is integrable on $I$ for $0 \le j \le k - 2$ (program's hypothesis (2) at order $k$);
$t \mapsto \partial^{k - 1} h / \partial x^{k - 1} (x, t) = \partial^{k - 1} f / \partial x^{k - 1} (x, t)$ is p.m. on $I$ (it is even integrable by hypothesis (2));
on $[a, b] \subset A$, $\lvert \partial^{k - 1} h / \partial x^{k - 1} (x, \cdot) \rvert \le \psi_{k - 1}$ integrable on $I$ (Step 1).

By the inductive hypothesis, $g = \int_I h(\cdot, t)\,\mathrm{d}t$ is of class $\mathcal{C}^{k - 1}$ on $[a, b]$ and $$ g^{(k - 1)}(x) = \int_I \frac{\partial^{k - 1} f}{\partial x^{k - 1}}(x, t)\,\mathrm{d}t \quad \text{for } x \in [a, b]. $$ Step 3 -- applying the $\mathcal{C}^1$ Leibniz theorem to $g^{(k - 1)}$. Define $\tilde{h}(x, t) = \partial^{k - 1} f / \partial x^{k - 1} (x, t)$. The hypotheses of the $\mathcal{C}^1$ Leibniz theorem (local form) for $\tilde{h}$ on $[a, b] \times I$:

$x \mapsto \tilde{h}(x, t)$ is $\mathcal{C}^1$ on $[a, b]$ (since $f$ is $\mathcal{C}^k$), with derivative $\partial^k f / \partial x^k$;
$t \mapsto \tilde{h}(x, t) = \partial^{k - 1} f / \partial x^{k - 1} (x, t)$ is integrable on $I$ for every $x \in [a, b]$ (program's hypothesis (2));
$t \mapsto \partial \tilde{h} / \partial x (x, t) = \partial^k f / \partial x^k (x, t)$ is p.m. on $I$ (program's hypothesis (3));
$\lvert \partial \tilde{h} / \partial x (x, \cdot) \rvert = \lvert \partial^k f / \partial x^k (x, \cdot) \rvert \le \varphi_{[a, b]}$ on $I$ (program's hypothesis (4)).

By Leibniz (§4.1, local form), $x \mapsto \int_I \tilde{h}(x, t)\,\mathrm{d}t = g^{(k - 1)}(x)$ is $\mathcal{C}^1$ on $[a, b]$ and its derivative is $\int_I \partial^k f / \partial x^k (x, t)\,\mathrm{d}t$. So $g$ is $\mathcal{C}^k$ on $[a, b]$ with $g^{(k)}(x) = \int_I \partial^k f / \partial x^k (x, t)\,\mathrm{d}t$. Since $[a, b] \subset A$ was arbitrary, $g$ is $\mathcal{C}^k$ on $A$.

Remark (why the $\mathcal{C}^k$ theorem is stated locally). The conclusion "$g \in \mathcal{C}^k(A)$" is a local property: $g$ is of class $\mathcal{C}^k$ on $A$ if and only if it is of class $\mathcal{C}^k$ on every segment $[a, b] \subset A$. This is why the natural statement is the local one, and why the proof works segment by segment --- the intermediate dominants $\psi_j$ are built on each fixed segment $[a, b]$, where the factors $(b - a)^{m - j}$ are finite and harmless. A single global $\varphi$ dominating $\partial^k f / \partial x^k$ over all of $A$ --- even an unbounded $A$ --- is also sufficient: it restricts, on every segment $[a, b] \subset A$, to a valid local dominant $\varphi_{[a, b]} = \varphi$, so the theorem applies on each segment and hence on $A$. The example below establishing the $\mathcal{C}^\infty$ regularity of $x \mapsto \int_0^{+\infty} e^{-t}\sin(xt)\,\mathrm{d}t$ on the unbounded $\mathbb{R}$ uses exactly such a single global dominant. In practice, however, one verifies the domination segment by segment, exactly as the program prescribes --- a global dominant is a bonus, never a requirement.

Example

Gamma function -- class $\mathcal{C}^\infty$. Show that $\Gamma$ is of class $\mathcal{C}^\infty$ on $]0, +\infty[$ with $\Gamma^{(k)}(x) = \int_0^{+\infty} (\ln t)^k\,e^{-t} t^{x - 1}\,\mathrm{d}t$ for every $k \ge 0$.
By induction, applying the $\mathcal{C}^k$ Leibniz theorem on every segment $[a, b] \subset ]0, +\infty[$. The $k$-th derivative in $x$ of $e^{-t} t^{x - 1} = e^{-t} e^{(x - 1)\ln t}$ is $(\ln t)^k\,e^{-t} t^{x - 1}$. Local domination on $[a, b]$: for $x \in [a, b]$ and $t > 0$, $$ \lvert (\ln t)^k\,e^{-t} t^{x - 1} \rvert \le \varphi_{[a, b], k}(t) := \begin{cases} (\lvert \ln t \rvert)^k\,e^{-t}\,t^{a - 1} & \text{if } 0 < t \le 1, \\ (\ln t)^k\,e^{-t}\,t^{b - 1} & \text{if } t \ge 1, \end{cases} $$ integrable on $]0, +\infty[$ by exponential decay (at $+\infty$) and Riemann-Bertrand comparison (at $0$, $(\lvert \ln t \rvert)^k\,t^{a - 1} = o(t^{a - 1 - \delta})$ for any $\delta > 0$, and the integral $\int_0^1 t^{a - 1 - \delta}\,\mathrm{d}t$ converges for $\delta < a$). The integrability hypothesis on the intermediate derivatives is met by the same Riemann-Bertrand argument with any $j < k$ in place of $k$: $t \mapsto (\ln t)^j\,e^{-t} t^{x - 1}$ is integrable on $]0, +\infty[$ for every $x > 0$. By Leibniz $\mathcal{C}^k$, $\Gamma$ is $\mathcal{C}^k$ on $[a, b]$ for every $k$. Since $k$ is arbitrary and $[a, b] \subset ]0, +\infty[$ is arbitrary, $\Gamma \in \mathcal{C}^\infty(]0, +\infty[)$.

Example

Continuing the starter: $g(x) = \displaystyle\int_0^{+\infty} e^{-t} \sin(xt)\,\mathrm{d}t$ is of class $\mathcal{C}^\infty$ on $\mathbb{R}$.
For every $t \ge 0$, $x \mapsto e^{-t} \sin(xt)$ is $\mathcal{C}^\infty$ on $\mathbb{R}$ with $\partial^k f / \partial x^k (x, t) = e^{-t}\,t^k\,\sin^{(k)}(xt)$ where $\sin^{(k)}$ is a sine or cosine (with sign), bounded by $1$. Global domination of $\partial^k f / \partial x^k$ on $\mathbb{R} \times [0, +\infty[$: $\lvert \partial^k f / \partial x^k (x, t) \rvert \le t^k\,e^{-t} =: \varphi_k(t)$, integrable on $[0, +\infty[$ with $\int_0^\infty t^k e^{-t}\,\mathrm{d}t = k!$. Integrability of the intermediate derivatives $\partial^j f / \partial x^j$ is checked the same way. By Leibniz $\mathcal{C}^k$ for every $k$, $g \in \mathcal{C}^\infty(\mathbb{R})$.

Skills to practice

Establishing $\mathcal{C}^k$ or $\mathcal{C}^\infty$ regularity

IV.3 Asymptotic study of a parameter integral

The program's capacity « exemples d'études de fonctions définies comme intégrales à paramètre : régularité, étude asymptotique » has two halves: regularity (§3 + §4.1 + §4.2 above) and asymptotic study. This last subsection collects the standard safe recipe --- change of variable + continuous CD --- and closes the chapter with one fully worked study, the equivalent of $\Gamma(x)$ as $x \to 0^+$.

Method — Asymptotic study via change of variable and continuous CD

The single safe recipe (the unsafe shortcut "find an equivalent of $f(x, t)$ inside the integral and integrate the equivalent" is illegitimate without a uniform domination):

identify the asymptotic direction ($x \to x_0$, possibly $\pm\infty$) and the integrand's dominant behaviour in that limit;
pick a normalising substitution $u = \alpha(x)\,t$ (or $u = \beta(x)(t - c(x))$) chosen so that $f(x, t)\,\mathrm{d}t = \beta(x)\,h(x, u)\,\mathrm{d}u$ with $h(x, u)$ admitting a simple limit $h_\infty(u)$ as $x \to x_0$ and a common integrable dominant $\varphi(u)$ on the resulting fixed interval $J$;
apply the continuous CD theorem (§1.2) to deduce $\int_J h(x, u)\,\mathrm{d}u \to L := \int_J h_\infty(u)\,\mathrm{d}u$;
provided $L \neq 0$, conclude the equivalent $g(x) \sim L\,\beta(x)$ as $x \to x_0$. (If $L = 0$, dominated convergence yields only $g(x) = o(\beta(x))$ --- a sharper analysis, e.g. a higher-order expansion of $h(x, u)$, is then needed to find the true equivalent.)

In favourable cases the substitution is the identity --- continuous CD applies directly to $f(x, t)$.

Example

Asymptotic of $\Gamma$ at $0$: show that $\Gamma(x) \sim 1/x$ as $x \to 0^+$.
The functional relation $\Gamma(x + 1) = x\,\Gamma(x)$ (recalled by IPP from the definition: $\Gamma(x + 1) = \int_0^\infty e^{-t} t^x\,\mathrm{d}t = [-e^{-t} t^x]_0^\infty + x \int_0^\infty e^{-t} t^{x - 1}\,\mathrm{d}t = x\,\Gamma(x)$, the bracket vanishing at both ends for $x > 0$) reduces the asymptotic to showing $\Gamma(x + 1) \to 1$ as $x \to 0^+$.
Apply continuous CD directly (no substitution needed) to $f_x(t) = e^{-t} t^x$ on $I = ]0, +\infty[$ with $x \in J = ]0, 1]$ and $x_0 = 0^+$:

each $f_x$ is continuous on $]0, +\infty[$;
for every $t > 0$, $f_x(t) = e^{-t} t^x = e^{-t}\,e^{x \ln t} \xrightarrow[x \to 0^+]{} e^{-t}$ (continuity of the exponential); the limit $t \mapsto e^{-t}$ is continuous on $]0, +\infty[$;
common domination on $]0, 1] \times ]0, +\infty[$: for $x \in ]0, 1]$ and $t > 0$, $e^{-t} t^x \le e^{-t} \max(1, t)$, an integrable dominant on $]0, +\infty[$ (decomposed as $e^{-t}$ on $]0, 1]$ and $t\,e^{-t}$ on $[1, +\infty[$, both integrable).

By continuous CD (§1.2), $$ \Gamma(x + 1) = \int_0^{+\infty} e^{-t} t^x\,\mathrm{d}t \xrightarrow[x \to 0^+]{} \int_0^{+\infty} e^{-t}\,\mathrm{d}t = 1. $$ Hence $\Gamma(x) = \Gamma(x + 1)/x \sim 1/x$ as $x \to 0^+$. This closes the chapter's $\Gamma$ thread: definition + continuity (§3.1), class $\mathcal{C}^1$ (§4.1), class $\mathcal{C}^\infty$ (§4.2), and asymptotic behaviour at $0$ (§4.3).

Skills to practice

Finding an asymptotic equivalent