CommeUnJeu · L2 MP

Integration on an arbitrary interval

⌚ ~83 min ▢ 10 blocks ✓ 18 exercises ➣ Prerequisites : Integration on a segment, Asymptotic analysis, Numerical and vector series

The MPSI integral was defined on a compact segment $[a, b]$ for piecewise continuous (CM) functions. This chapter extends the construction to non-compact intervals --- $[a, +\infty[$, $]a, b]$, $]a, b[$ --- by passage \`a la limite at one or two endpoints. Two new tools structure the chapter. First, integrability: a sufficient condition that mirrors absolute convergence for series, with $L^1(I, \mathbb{K})$ as the $\mathbb{K}$-vector space of integrable functions of $I$. Second, the integration of comparison relations: the integral mirror of the summation of comparison relations from Numerical and vector series, with the same structure (positive reference; transfer to remainders if integrable, to partial integrals if not).
The chapter is calibrated to support an asymptotic toolbox: extracting equivalents and asymptotic developments of integrals as $x \to +\infty$ or near a singularity, with the Gaussian asymptotic $\int_x^{+\infty} \mathrm{e}^{-t^2}\,\mathrm{d}t \;\sim\; \mathrm{e}^{-x^2}/(2x)$ as the headline application.

Conventions

Throughout this chapter, $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$, and $I$ is an interval of $\mathbb{R}$ of non-empty interior. We write $\mathcal{C}_{pm}(I, \mathbb{K})$ for the $\mathbb{K}$-vector space of piecewise continuous functions of $I$ --- the same construction as MPSI's $\mathcal{CM}([a, b], \mathbb{K})$ in Integration on a segment, now applied to an arbitrary interval $I$ instead of a compact segment (a function $f : I \to \mathbb{K}$ is CM on $I$ if its restriction to every compact sub-segment of $I$ is CM in the MPSI sense). The « pm » (« piecewise » / « par morceaux ») subscript is the prépa convention used throughout this chapter. Improper integral notation: $\int_a^{+\infty} f$, $\int_a^b f$ with $b$ open (or $\int_a^b f(t)\,\mathrm{d}t$), $\int_I f$ for an integrable $f$.

I Improper integrals

I.1 Improper integral on a half-line at infinity

From the segment integral of Integration on a segment, we extend by passing to the limit at the upper endpoint $+\infty$. The function $f$ stays in the same class --- piecewise continuous --- only the endpoint changes.

Definition — Improper integral on a half-line at infinity

Let $f \in \mathcal{C}_{pm}([a, +\infty[, \mathbb{K})$. The improper integral of $f$ on $[a, +\infty[$ is said to converge when the function $x \mapsto \int_a^x f$ admits a finite limit as $x \to +\infty$. The limit is then denoted $$ \int_a^{+\infty} f \;=\; \lim_{x \to +\infty} \int_a^x f \;\in\; \mathbb{K}, $$ or equivalently $\int_a^{+\infty} f(t)\,\mathrm{d}t$. Otherwise the integral is said to diverge. When $f$ is real-valued and $x \mapsto \int_a^x f$ tends to $\pm\infty$, one writes by convention $\int_a^{+\infty} f = \pm\infty$.

Example — Riemann integral at infinity converges

The primitive of $t \mapsto 1/t^2$ on $[1, +\infty[$ is $t \mapsto -1/t$, so $$ \int_1^x \frac{\mathrm{d}t}{t^2} \;=\; \left[-\frac{1}{t}\right]_1^x \;=\; 1 - \frac{1}{x} \;\xrightarrow[x \to +\infty]{}\; 1. $$ The improper integral converges, with $\int_1^{+\infty} \mathrm{d}t/t^2 = 1$.

Example — Harmonic integral diverges to infinity

A primitive of $1/t$ on $[1, +\infty[$ is $\ln$, so $\int_1^x \mathrm{d}t/t = \ln x \to +\infty$. By convention we write $\int_1^{+\infty} \mathrm{d}t/t = +\infty$.

Example — Cosine integral diverges by oscillation

$\int_0^x \cos t\,\mathrm{d}t = \sin x$, which has no limit as $x \to +\infty$ (oscillates between $-1$ and $1$). The integral diverges --- and this is not a divergence to $\pm\infty$.

Proposition — Derivative of a tail integral

If $f \in \mathcal{C}([a, +\infty[, \mathbb{K})$ and $\int_a^{+\infty} f$ converges, the function $$ G : x \;\longmapsto\; \int_x^{+\infty} f \;=\; \int_a^{+\infty} f - \int_a^x f $$ is differentiable on $[a, +\infty[$ with $G'(x) = -f(x)$ (the standard convention treats the right derivative at $a$ as the derivative there).

Proof

By the fundamental theorem of calculus (recalled from Integration on a segment) applied to the continuous function $f$ on $[a, +\infty[$, the map $x \mapsto \int_a^x f$ is differentiable on $[a, +\infty[$ with derivative $f$. Subtracting from the constant $\int_a^{+\infty} f$ reverses the sign.

Proposition — Divergence grossi\`ere \`a l'infini

Let $f \in \mathcal{C}_{pm}([a, +\infty[, \mathbb{R})$. If $f$ admits a non-zero finite limit $\ell$ or $\pm\infty$ at $+\infty$, then $\int_a^{+\infty} f$ diverges.

Proof

The argument rests on the eventual constant sign of $f$:

If $f(x) \to \ell > 0$, then from some rank $N$, $f(x) \ge \ell/2 > 0$ for $x \ge N$, so $\int_N^x f \ge (\ell/2)(x - N) \to +\infty$, hence $\int_a^x f \to +\infty$.
The case $\ell < 0$ is symmetric, with $\int_a^x f \to -\infty$.
The case $\ell = +\infty$ is even more immediate: from some rank, $f(x) \ge 1$, so $\int_a^x f \ge x - a - C \to +\infty$. The case $\ell = -\infty$ is symmetric.

The eventual constant sign of $f$ is what drives the integral to $\pm\infty$.

Caution

Unlike the series case (where $u_n \to 0$ is necessary for convergence of $\sum u_n$), the integral $\int_a^{+\infty} f$ can converge without $f$ admitting a limit at $+\infty$ at all. The divergence-grossi\`ere proposition is a one-way sufficient condition for divergence, not the contrapositive of an « $f \to 0$ necessary » statement.
Counterexample. A positive « triangular » function $f$ on $[2, +\infty[$: triangles of base $1/n^2$, height $2$, centred on each integer $n \ge 3$ (so each triangle stays strictly inside $[2, +\infty[$; $f = 0$ on $[2, 3 - 1/18]$). The triangle at $n$ has area $1/n^2$, so $\int_2^{x} f \le \sum_{k \ge 3} 1/k^2 \le \pi^2/6 - 1 - 1/4$, and $\int_2^{+\infty} f$ converges. Yet $f$ reaches the value $2$ at every integer $n \ge 3$, so $f$ has no limit at $+\infty$.

Method — Checking convergence of an improper integral at infinity

Confirm $f \in \mathcal{C}_{pm}([a, +\infty[, \mathbb{K})$.
If $f$ has a non-zero limit (finite or $\pm\infty$) at $+\infty$: divergence by divergence-grossi\`ere.
Otherwise, try to compute a primitive of $f$ on $[a, +\infty[$ and pass to the limit.
If no closed primitive: fall back on the positive-case characterisation of \S 2.1 or the integrability comparison theorem of \S 2.3.

Do not mimic the series rule « $u_n \to 0$ necessary » --- the triangular counterexample above shows this rule is false for integrals.

Skills to practice

Computing an improper integral on a half-line at infinity

I.2 Improper integral on an arbitrary interval

The construction of \S 1.1 extends to the other two interval shapes by symmetry: semi-open $]a, b]$ (pass to the limit at $a^+$), open $]a, b[$ (pass to the limit at both endpoints, independently). We also isolate the « faussement impropre » case where a finite limit at the open endpoint makes the integral genuinely a segment integral in disguise.

Definition — Improper integral on a left half-open interval

Let $f \in \mathcal{C}_{pm}(]a, b], \mathbb{K})$ (with $a \in \mathbb{R} \cup \{-\infty\}$, $b \in \mathbb{R}$). The improper integral $\int_a^b f$ converges when the function $x \mapsto \int_x^b f$ admits a finite limit as $x$ tends to the open endpoint $a$ from inside $]a, b]$ --- that is, $x \to a^+$ if $a \in \mathbb{R}$, or $x \to -\infty$ if $a = -\infty$. The limit is then $$ \int_a^b f \;=\; \lim_{x \to a} \int_x^b f. $$

Example — A left half-open improper integral by primitive

For $f(t) = 1/\sqrt{t}$ on $]0, 1]$, $f$ is continuous on $]0, 1]$ and a primitive is $t \mapsto 2\sqrt{t}$, so $$ \int_x^1 \frac{\mathrm{d}t}{\sqrt{t}} \;=\; \bigl[2\sqrt{t}\bigr]_x^1 \;=\; 2 - 2\sqrt{x} \;\xrightarrow[x \to 0^+]{}\; 2. $$ The improper integral $\int_0^1 \mathrm{d}t/\sqrt{t}$ converges, with value $2$. (Same conclusion via the §1.4 reference: $\int_0^1 \mathrm{d}t/t^\alpha$ converges for $\alpha = 1/2 < 1$.)

Definition — Improper integral on an open interval

Let $f \in \mathcal{C}_{pm}(]a, b[, \mathbb{K})$ (with $a \in \mathbb{R} \cup \{-\infty\}$, $b \in \mathbb{R} \cup \{+\infty\}$). Pick any $c \in ]a, b[$. The improper integral $\int_a^b f$ converges when both $\int_a^c f$ (improper at $a$) and $\int_c^b f$ (improper at $b$) converge, and then $$ \int_a^b f \;:=\; \int_a^c f \,+\, \int_c^b f. $$ The value does not depend on $c$ (consequence of Chasles on segments). The two convergences are required independently: it is not enough that a symmetric / principal-value limit (e.g. $\lim_{r \to +\infty} \int_{-r}^{r} f$ for $]-\infty, +\infty[$) exists --- see the Example below.

Example — Symmetric limit is not the same as convergence

The integral $\int_{-\infty}^{+\infty} t\,\mathrm{d}t$ diverges: indeed, $\int_0^x t\,\mathrm{d}t = x^2/2 \to +\infty$, so $\int_0^{+\infty} t\,\mathrm{d}t$ diverges, and the open-interval definition fails at $+\infty$. Yet by symmetry, $\int_{-x}^{x} t\,\mathrm{d}t = 0$ for every $x > 0$, so the symmetric limit is $0$ --- a misleading « cancellation » that is not the definition of convergence on $]-\infty, +\infty[$.

Proposition — Faussement impropre integral

Let $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$ with $b \in \mathbb{R}$ finite. Suppose $f$ is continuous on a left-neighbourhood of $b$ and admits a finite limit at $b^-$. Define the extension $\tilde{f}$ on $[a, b]$ by $\tilde{f}(b) := \lim_{x \to b^-} f(x)$ (and $\tilde{f} = f$ on $[a, b[$). Then $\tilde{f} \in \mathcal{C}_{pm}([a, b], \mathbb{K})$, the improper integral $\int_a^b f$ converges, and $$ \int_a^b f \;=\; \int_a^b \tilde{f}, $$ the latter being a segment integral.

Proof

The left-neighbourhood-continuity hypothesis prevents CM jumps from accumulating at $b$, so $\tilde{f}$ has only the finitely many interior jumps of $f$ on $[a, b[$, plus continuity at $b$ by construction --- thus $\tilde{f} \in \mathcal{C}_{pm}([a, b], \mathbb{K})$. The map $x \mapsto \int_a^x \tilde{f}$ is continuous on $[a, b]$ (recalled from Integration on a segment); on $[a, b[$ it coincides with $x \mapsto \int_a^x f$, so the latter has at $b^-$ the limit $\int_a^b \tilde{f}$.

Example — Sine cardinal faussement impropre at zero

The function $t \mapsto (\sin t)/t$ is continuous on $]0, 1]$ and $(\sin t)/t \to 1$ as $t \to 0^+$ (Taylor expansion $\sin t = t + O(t^3)$). By the proposition (applied symmetrically at the left endpoint $0$, per the chapter-wide convention of \S 1.2): $\int_0^1 (\sin t)/t\,\mathrm{d}t$ converges as a faussement impropre integral, equal to the segment integral of the continuous extension.

Example — Logarithm integral genuinely improper at zero

The function $\ln$ does not have a finite limit at $0^+$ ($\ln t \to -\infty$), so the proposition does not apply. The integral is nonetheless convergent by direct computation: a primitive of $\ln$ is $t \mapsto t \ln t - t$, so $$ \int_x^1 \ln t\,\mathrm{d}t \;=\; [t \ln t - t]_x^1 \;=\; -1 - (x \ln x - x) \;\xrightarrow[x \to 0^+]{}\; -1, $$ using $x \ln x \to 0$.

Skills to practice

Studying an improper integral on an arbitrary interval

I.3 Properties and notations

The four properties below are direct consequences of the corresponding properties of the segment integral (recalled from Integration on a segment) plus passage à la limite. Each is stated for the case $[a, b[$ with $b \in \mathbb{R} \cup \{+\infty\}$ --- the other interval shapes are identical by symmetry.

Proposition — Linearity

Let $f, g \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$ with $\int_a^b f$ and $\int_a^b g$ convergent, and $\lambda, \mu \in \mathbb{K}$. Then $\int_a^b (\lambda f + \mu g)$ converges and $$ \int_a^b (\lambda f + \mu g) \;=\; \lambda \int_a^b f \;+\; \mu \int_a^b g. $$ The set of $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$ with $\int_a^b f$ convergent is a $\mathbb{K}$-subspace, and the integral is linear on it.

Proof

Linearity of the segment integral (recalled from Integration on a segment) gives, for every $x \in [a, b[$, $$ \int_a^x (\lambda f + \mu g) \;=\; \lambda \int_a^x f \,+\, \mu \int_a^x g. $$ Pass to the limit as $x \to b^-$: each term on the right has a finite limit, so the left-hand side does too.

Proposition — Chasles

Let $f \in \mathcal{C}_{pm}(]a, b[, \mathbb{K})$ with $\int_a^b f$ convergent, and $c \in ]a, b[$. Then $\int_a^c f$ and $\int_c^b f$ both converge, and $$ \int_a^b f \;=\; \int_a^c f \;+\; \int_c^b f. $$

Proof

Direct from the definition of convergence on the open interval, which is precisely the simultaneous convergence of $\int_a^c f$ and $\int_c^b f$ for any $c \in ]a, b[$, plus segment Chasles for the equality.

Proposition — Positivity and monotonicity (real case)

For real-valued $f \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ with $\int_a^b f$ convergent:

if $f \ge 0$ on $[a, b[$, then $\int_a^b f \ge 0$;
if $f, g \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ with $f \le g$ and both $\int_a^b f$, $\int_a^b g$ converge, then $\int_a^b f \le \int_a^b g$.

Proof

Limit of positive (resp. ordered) segment integrals. The order between limits of real sequences is preserved by passage à la limite.

Proposition — Real-imaginary split for complex-valued integrals

Let $f \in \mathcal{C}_{pm}(]a, b[, \mathbb{C})$. Then $\int_a^b f$ converges if and only if both $\int_a^b \operatorname{Re}(f)$ and $\int_a^b \operatorname{Im}(f)$ converge, and in case of convergence $$ \int_a^b f \;=\; \int_a^b \operatorname{Re}(f) \;+\; i \int_a^b \operatorname{Im}(f). $$

Proof

A complex sequence converges in $\mathbb{C}$ if and only if its real and imaginary parts converge in $\mathbb{R}$. Apply this to the segment integral $y \mapsto \int_c^y f$ (with $\operatorname{Re}(\int_c^y f) = \int_c^y \operatorname{Re}(f)$ by linearity of the segment integral) for any interior $c \in ]a, b[$: the upper-endpoint improper integral $\int_c^b f$ converges iff $\int_c^b \operatorname{Re}(f)$ and $\int_c^b \operatorname{Im}(f)$ converge. Symmetrically for the lower-endpoint half-integral on $]a, c]$. Combining the two halves by linearity yields the announced equivalence for $\int_a^b f$.

Definition — Divergence-to-infinity notation for positive integrands

For $f \in \mathcal{C}_{pm}(I, \mathbb{R}^+)$ with $\int_I f$ divergent, one writes by convention $\int_I f = +\infty$. As a practical capacité (program § 8a): a computation yielding $\int_I f < +\infty$ for a positive-valued $f$ counts as a proof of convergence.

Example — The notation in action

By this convention, $\int_1^{+\infty} \mathrm{d}t/t = +\infty$ (recall Example § 1.1: the integrand is positive on $[1, +\infty[$, $\int_1^x \mathrm{d}t/t = \ln x \to +\infty$). Conversely, computing $\int_1^x \mathrm{d}t/t^2 = 1 - 1/x \le 1$ for every $x \ge 1$ proves $\int_1^{+\infty} \mathrm{d}t/t^2 < +\infty$, hence convergence (no need to identify the limit value $1$ explicitly).

Skills to practice

Applying linearity$\virgule$ Chasles and the divergence notation

I.4 Reference improper integrals

Four computed improper integrals recur throughout asymptotic analysis and are to be known by heart. Each is settled by an explicit primitive plus passage à la limite.

Proposition — Riemann integrals at infinity

For $\alpha \in \mathbb{R}$, $$ \int_1^{+\infty} \frac{\mathrm{d}t}{t^\alpha} \quad\text{converges} \iff \alpha > 1. $$ In that case the value is $1/(\alpha - 1)$.

Proof

For $\alpha \ne 1$, the primitive of $t \mapsto 1/t^\alpha$ is $t \mapsto t^{1 - \alpha}/(1 - \alpha)$, so $$ \int_1^x \frac{\mathrm{d}t}{t^\alpha} \;=\; \frac{x^{1 - \alpha} - 1}{1 - \alpha} \;=\; \frac{1}{\alpha - 1}\bigl(1 - x^{1 - \alpha}\bigr). $$ As $x \to +\infty$, $x^{1 - \alpha} \to 0$ if $\alpha > 1$ (finite limit $1/(\alpha - 1)$) and $x^{1 - \alpha} \to +\infty$ if $\alpha < 1$ (divergence).
For $\alpha = 1$, $\int_1^x \mathrm{d}t/t = \ln x \to +\infty$. Divergence.

Proposition — Riemann integrals at zero

For $\alpha \in \mathbb{R}$, $$ \int_0^1 \frac{\mathrm{d}t}{t^\alpha} \quad\text{converges} \iff \alpha < 1. $$

Proof

For $\alpha \ne 1$, the primitive $t \mapsto t^{1 - \alpha}/(1 - \alpha)$ gives $\int_x^1 \mathrm{d}t/t^\alpha = (1 - x^{1 - \alpha})/(1 - \alpha)$. As $x \to 0^+$: $x^{1 - \alpha} \to 0$ if $\alpha < 1$ (finite limit $1/(1 - \alpha)$), $x^{1 - \alpha} \to +\infty$ if $\alpha > 1$ (divergence).
For $\alpha = 1$, the primitive is $\ln$ (not the power $t^{1-\alpha}/(1-\alpha)$, which is invalid at this value), so $\int_x^1 \mathrm{d}t/t = -\ln x \to +\infty$. Divergence.

Proposition — Exponential at infinity

For $a \in \mathbb{R}$:

if $a > 0$, $\int_0^{+\infty} \mathrm{e}^{-at}\,\mathrm{d}t$ converges with value $1/a$;
if $a = 0$, $\int_0^{+\infty} 1\,\mathrm{d}t = +\infty$ (trivial divergence);
if $a < 0$, the integral diverges by divergence-grossière ($\mathrm{e}^{-at} \to +\infty$).

Proof

For $a > 0$, the primitive of $\mathrm{e}^{-at}$ is $-\mathrm{e}^{-at}/a$, so $$ \int_0^x \mathrm{e}^{-at}\,\mathrm{d}t \;=\; \left[-\frac{\mathrm{e}^{-at}}{a}\right]_0^x \;=\; \frac{1 - \mathrm{e}^{-ax}}{a} \;\xrightarrow[x \to +\infty]{}\; \frac{1}{a}. $$ For $a \le 0$, the integrand does not tend to $0$ at $+\infty$ (it tends to $1$ if $a = 0$, to $+\infty$ if $a < 0$), so divergence-grossière of \S 1.1 forces divergence.

Proposition — Logarithm at zero

$\int_0^1 \ln t\,\mathrm{d}t$ converges, with value $-1$.

Proof

A primitive of $\ln$ is $t \mapsto t \ln t - t$, so $\int_x^1 \ln t\,\mathrm{d}t = -1 - (x \ln x - x) \to -1$ as $x \to 0^+$ (using $x \ln x \to 0$).

Method — Recognising a reference integral

Read off the parameter from the integrand --- $\alpha$ for a Riemann integral, $a$ for an exponential --- and check the threshold ($\alpha > 1$ at $+\infty$, $\alpha < 1$ at $0$, $a > 0$ for the exponential). Memorise the four references; the comparison theorems of § 2.3 then reduce most integrability questions in this chapter to comparing the integrand to one of these four references.

Skills to practice

Identifying a reference improper integral

I.5 IPP and change of variable on an arbitrary interval

Both MPSI segment theorems --- integration by parts and change of variable --- extend to the non-compact case under the same regularity hypotheses, plus the existence of finite limits at the open endpoints. In practical applications, one does not repeat the regularity hypotheses (program's stated permission).

Theorem — IPP on an arbitrary interval

Let $f, g \in \mathcal{C}^1([a, b[, \mathbb{K})$ with $b \in \mathbb{R} \cup \{+\infty\}$. If $\lim_{x \to b^-} f(x) g(x)$ exists and is finite, then the improper integrals $\int_a^b f' g$ and $\int_a^b f g'$ are of the same nature, and in case of convergence $$ \int_a^b f g' \;=\; \bigl[fg\bigr]_a^{b^-} \,-\, \int_a^b f' g, \qquad \bigl[fg\bigr]_a^{b^-} \,:=\, \lim_{x \to b^-} f(x)g(x) \,-\, f(a)g(a). $$ Bilateral version on $]a, b[$: if both $\lim_{x \to a^+} f(x) g(x)$ and $\lim_{x \to b^-} f(x) g(x)$ exist and are finite, the same conclusion holds with $\bigl[fg\bigr]_{a^+}^{b^-} = \lim_{b^-} fg - \lim_{a^+} fg$.

Proof

IPP on the segment $[a, x]$ (recalled from Integration on a segment): $$ \int_a^x f g' \;=\; [fg]_a^x \,-\, \int_a^x f' g. $$ As $x \to b^-$: the boundary term $[fg]_a^x$ tends to $\lim_{b^-} fg - f(a)g(a)$, finite by hypothesis. So $\int_a^x fg'$ and $\int_a^x f'g$ differ by a quantity with a finite limit; they have the same nature, and the announced equality follows. The bilateral case follows by splitting at any interior point and applying the unilateral version on each side.

Method — Applying IPP on an arbitrary interval

Compute the boundary term $[fg]$ first: it must have finite limits at the open endpoints. If so, IPP transforms the integral; if not, IPP does not apply directly --- one must first dampen $f$ or $g$ (e.g. integrate by parts on a slightly smaller interval first).

Example — First exponential moment via IPP

Set $u(t) = t$, $v'(t) = \mathrm{e}^{-t}$, so $u'(t) = 1$, $v(t) = -\mathrm{e}^{-t}$. The boundary term $u v = -t \mathrm{e}^{-t}$ has limits $0$ at $0$ and $0$ at $+\infty$ (the second by $t \mathrm{e}^{-t} \to 0$); finite, so IPP applies. Then $$ \int_0^{+\infty} t \mathrm{e}^{-t}\,\mathrm{d}t \;=\; \bigl[-t \mathrm{e}^{-t}\bigr]_0^{+\infty} \,+\, \int_0^{+\infty} \mathrm{e}^{-t}\,\mathrm{d}t \;=\; 0 \,+\, 1 \;=\; 1. $$

Theorem — Change of variable on an arbitrary interval

Let $f$ be continuous on $]a, b[$ and $\varphi : ]\alpha, \beta[ \,\to\, ]a, b[$ a bijective, strictly increasing, $\mathcal{C}^1$ map. Then the integrals $$ \int_a^b f(t)\,\mathrm{d}t \quad\text{and}\quad \int_\alpha^\beta f\bigl(\varphi(u)\bigr) \varphi'(u)\,\mathrm{d}u $$ are of the same nature, and equal in case of convergence. If $\varphi$ is strictly decreasing bijective $\mathcal{C}^1$, the same conclusion holds: now $\varphi(\alpha^+) = b^-$ and $\varphi(\beta^-) = a^+$, so $\int_a^b f(t)\,\mathrm{d}t = \int_\beta^\alpha f(\varphi(u)) \varphi'(u)\,\mathrm{d}u = -\int_\alpha^\beta f(\varphi(u)) \varphi'(u)\,\mathrm{d}u$ (the bounds reverse automatically, then one flips them back at the cost of the leading sign).

Proof

Fix any $\gamma \in ]\alpha, \beta[$, and let $c := \varphi(\gamma) \in ]a, b[$. By the open-interval definition (\S 1.2), $\int_a^b f$ converges iff $\int_a^c f$ and $\int_c^b f$ converge independently; same for the right-hand side at $\gamma$. By the change-of-variable theorem on a segment (recalled from Integration on a segment), for every segment $[\gamma, \beta'] \subset [\gamma, \beta[$: $$ \int_c^{\varphi(\beta')} f(t)\,\mathrm{d}t \;=\; \int_\gamma^{\beta'} f\bigl(\varphi(u)\bigr) \varphi'(u)\,\mathrm{d}u. $$ Pass to the limit $\beta' \to \beta^-$: by bijectivity and monotonicity of $\varphi$, $\varphi(\beta') \to b^-$, so the two upper-endpoint improper integrals have the same nature and the same value when convergent. Symmetrically for the lower-endpoint half-integrals (limit $\alpha' \to \alpha^+$, $\varphi(\alpha') \to a^+$). Combining the two halves by Chasles establishes the equivalence for the full open interval.

Method — Applying change of variable on a non-compact interval

Use the monotonicity of $\varphi$ to send the endpoints correctly. In standard applications, the $\mathcal{C}^1$ and bijectivity hypotheses are not repeated (program's permission). Common moves: $t = \tan u$ to send $]0, \pi/2[ \to ]0, +\infty[$; $t = \mathrm{e}^u$ to send $]-\infty, +\infty[ \to ]0, +\infty[$; $t = u^2$ to handle $1/\sqrt{t}$-type singularities.

Example — Arctangent integral via change of variable

Apply $t = \tan u$, with $\varphi : ]0, \pi/2[ \,\to\, ]0, +\infty[$ bijective strictly increasing $\mathcal{C}^1$, $\varphi'(u) = 1 + \tan^2 u = 1 + t^2$. Then $$ \int_0^{+\infty} \frac{\mathrm{d}t}{1 + t^2} \;=\; \int_0^{\pi/2} \frac{1 + \tan^2 u}{1 + \tan^2 u}\,\mathrm{d}u \;=\; \int_0^{\pi/2} \mathrm{d}u \;=\; \frac{\pi}{2}. $$

Skills to practice

Applying IPP and change of variable on an arbitrary interval

II Positive case and integrable functions

II.1 Positive case

For a positive-valued CM function, the convergence problem reduces to boundedness of the partial integrals --- exactly the same simplification as for positive-term series in Numerical and vector series \S 2.

Theorem — Positive-case characterisation

Let $f \in \mathcal{C}_{pm}([a, b[, \mathbb{R}^+)$. Then $\int_a^b f$ converges if and only if the map $x \mapsto \int_a^x f$ is bounded on $[a, b[$.

Proof

The map $F : x \mapsto \int_a^x f$ is increasing on $[a, b[$ --- for $x \le y$ in $[a, b[$, $F(y) - F(x) = \int_x^y f \ge 0$ by positivity of $f$ (segment-integral positivity, recalled from Integration on a segment; the differentiability of $F$ at jump points is irrelevant). The monotone limit theorem then states: $F$ has a finite limit at $b^-$ if and only if $F$ is bounded above on $[a, b[$. For an arbitrary interval $]a, b[$, apply this to both halves separately after splitting at any $c \in ]a, b[$.

Method — Convergence via bounded partial integrals

For a positive-valued $f$, a computation aboutissant à $\int_a^x f \le M < +\infty$ for every $x \in [a, b[$ suffices to prove convergence (program's stated permission). One does not need the explicit limit value.

Theorem — Majoration rule

Let $f, g \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ with $0 \le f \le g$ on $[a, b[$. If $\int_a^b g$ converges, then $\int_a^b f$ converges, and $\int_a^b f \le \int_a^b g$. Contrapositively, if $\int_a^b f$ diverges, then $\int_a^b g$ diverges.

Proof

For every $x \in [a, b[$, $\int_a^x f \le \int_a^x g \le \int_a^b g$ (the last step using convergence and positivity of $g$). So $x \mapsto \int_a^x f$ is bounded above by $\int_a^b g$. By the positive-case characterisation, $\int_a^b f$ converges, with $\int_a^b f \le \int_a^b g$ (passage to the limit in the inequality).

Example — Sine squared over polynomial converges

The integrand is positive and $\sin^2(t)/(1+t^2) \le 1/(1+t^2)$ for every $t \ge 0$. Since $\int_0^{+\infty} \mathrm{d}t/(1+t^2) = \pi/2$ (computed in § 1.5), the majoration rule applies: the integral converges.

Method — Choosing a majorant

Aim for a known reference integral (\S 1.4): Riemann $1/t^\alpha$ at $+\infty$ or $0$, exponential $\mathrm{e}^{-at}$ at $+\infty$, $|\ln t|$ at $0$ (the positive integrable reference, since $\ln t < 0$ on $]0, 1[$). The asymptotic behaviour of $f$ at the open endpoint dictates the reference to compare to.

Skills to practice

Exploiting the positive-case characterisation

II.2 Absolute convergence and the space of integrable functions

The integral analog of absolute convergence for series. The space $L^1(I, \mathbb{K})$ of integrable functions is the chapter's central object, governed by the triangle inequality.

Definition — Absolute convergence

Let $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$. The improper integral $\int_a^b f$ is said to be absolutely convergent when the positive numerical integral $\int_a^b |f|$ converges. Same definition on $]a, b]$ and $]a, b[$.

Example — A first absolutely convergent integral

For $t \ge 0$, $|\cos(t)/(1+t^2)| \le 1/(1+t^2)$, and $\int_0^{+\infty} \mathrm{d}t/(1+t^2) = \pi/2$ converges (§ 1.5 example). By the majoration rule applied to the positive integrand $|\cos(t)/(1+t^2)|$, $\int_0^{+\infty} |\cos(t)/(1+t^2)|\,\mathrm{d}t$ converges, so $\int_0^{+\infty} \cos(t)/(1+t^2)\,\mathrm{d}t$ is absolutely convergent.

Theorem — Absolute convergence implies convergence

If $\int_a^b f$ is absolutely convergent, then it is convergent.

Proof

Real case. Take $f \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ with $\int |f|$ convergent. The decomposition $0 \le f + |f| \le 2|f|$ shows that $f + |f|$ is positive-valued with $\int (f + |f|) \le 2 \int |f|$, hence convergent by the majoration rule. Since $\int |f|$ converges by hypothesis, linearity of convergent improper integrals (\S 1.3) gives $$ \int_a^b f \;=\; \int_a^b (f + |f|) \,-\, \int_a^b |f| \;\in\; \mathbb{R}, $$ a difference of two convergent improper integrals --- so $\int_a^b f$ converges.
Complex case. Take $f \in \mathcal{C}_{pm}([a, b[, \mathbb{C})$ with $\int |f|$ convergent. Then $|\!\operatorname{Re}(f)| \le |f|$ and $|\!\operatorname{Im}(f)| \le |f|$, so by the majoration rule $\int |\!\operatorname{Re}(f)|$ and $\int |\!\operatorname{Im}(f)|$ converge. By the real case applied to $\operatorname{Re}(f)$ and $\operatorname{Im}(f)$, $\int \operatorname{Re}(f)$ and $\int \operatorname{Im}(f)$ converge. By the real-imaginary split of \S 1.3, $\int f = \int \operatorname{Re}(f) + i \int \operatorname{Im}(f)$ converges.
No appeal to a Cauchy criterion is needed.

Remark -- semi-convergent integrals not a program objective

L'étude des intégrales semi-convergentes n'est pas un objectif du programme. (Program § 8b, reproduced verbatim.) The MP chapter therefore handles convergence almost exclusively via absolute convergence; the parallel statement appears in Numerical and vector series § 1.4 for semi-convergent series, where the same out-of-scope marker applies.

Definition — Integrable function and the space of integrable functions

Let $I$ be an interval of $\mathbb{R}$ and $f : I \to \mathbb{K}$ a function. $f$ is integrable on $I$ when $f \in \mathcal{C}_{pm}(I, \mathbb{K})$ and the improper integral $\int_I |f|$ converges. The set of integrable functions on $I$ to $\mathbb{K}$ is denoted $L^1(I, \mathbb{K})$. For $f \in L^1(I, \mathbb{K})$, the integral $\int_I f$ is defined and is automatically convergent (by absolute convergence).

Example — Componentwise integration of a matrix-valued function

The framework $L^1(I, \mathbb{K})$ above is defined only for $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, but extends naturally to a matrix-valued $f : I \to \mathcal{M}_n(\mathbb{K})$ by componentwise definition: $f$ is integrable on $I$ when each scalar entry $f_{ij} : I \to \mathbb{K}$ is integrable, and then $\int_I f := (\int_I f_{ij})_{1 \le i, j \le n}$ (matrix entry-by-entry). Concretely, for $f : [0, +\infty[ \,\to\, \mathcal{M}_n(\mathbb{R})$ defined by $f_{ij}(t) = a_{ij}\, \mathrm{e}^{-t}$ (with $a_{ij} \in \mathbb{R}$), each entry $|a_{ij}| \mathrm{e}^{-t}$ is integrable by majoration by the exponential reference of \S 1.4, so $f$ is integrable componentwise and $$ \int_0^{+\infty} f \;=\; \bigl(a_{ij}\bigr)_{1 \le i, j \le n}. $$ This componentwise construction mirrors the coordinate criterion for matrix series of Numerical and vector series \S 1.3.

Proposition — The space of integrable functions is a vector space

$L^1(I, \mathbb{K})$ is a $\mathbb{K}$-vector subspace of $\mathcal{C}_{pm}(I, \mathbb{K})$.

Proof

The zero function is integrable. For $\lambda \in \mathbb{K}$ and $f, g \in L^1(I, \mathbb{K})$, $|\lambda f + g| \le |\lambda||f| + |g|$. The right-hand side is the positive combination of two integrable functions, hence integrable by linearity + majoration; so $\int |\lambda f + g|$ converges. Therefore $\lambda f + g \in L^1(I, \mathbb{K})$.

Proposition — Triangle inequality

For $f \in L^1(I, \mathbb{K})$, $$ \left|\int_I f\right| \;\le\; \int_I |f|. $$

Proof

On any sub-segment $[c, d] \subset I$, the triangle inequality for the segment integral (recalled from Integration on a segment) gives $|\int_c^d f| \le \int_c^d |f|$. The limiting process depends on the type of $I$: for $I = [a, b[$, fix $c = a$ and let $d \to b^-$; for $I = ]a, b]$, fix $d = b$ and let $c \to a^+$; for $I = ]a, b[$, split at any interior point and apply the unilateral case to each half (each half-integral has only one open endpoint). In all cases, the left-hand side tends to $|\int_I f|$ by continuity of $|\cdot|$, the right-hand side to $\int_I |f|$, and the inequality is preserved at the limit.

Proposition — Continuous positive integrable + zero integral implies zero

Let $f : I \to \mathbb{R}^+$ be continuous on $I$ and integrable on $I$, with $\int_I f = 0$. Then $f \equiv 0$ on $I$.

Proof

For any segment $[c, d] \subset I$, $0 \le \int_c^d f \le \int_I f = 0$ (positivity + total integral $= 0$), so $\int_c^d f = 0$. By the MPSI segment version ($f \in \mathcal{C}([c, d], \mathbb{R}^+)$ with $\int_c^d f = 0$ implies $f \equiv 0$ on $[c, d]$, recalled from Integration on a segment), $f \equiv 0$ on $[c, d]$. As $[c, d]$ is arbitrary, $f \equiv 0$ on $I$.

Caution

Continuity is essential in the previous Proposition: a positive CM function with isolated positive values can have zero integral. The hypothesis « continuous » cannot be relaxed to « piecewise continuous ».

Example — Sine over square absolutely convergent at infinity

For $t \ge 1$, $|\sin(t)/t^2| \le 1/t^2$, and $\int_1^{+\infty} \mathrm{d}t/t^2$ is the Riemann reference with $\alpha = 2 > 1$, hence convergent. So $\int_1^{+\infty} \sin(t)/t^2\,\mathrm{d}t$ is absolutely convergent. (Lower bound $1$ chosen to avoid the singularity of $\sin(t)/t^2 \sim 1/t$ at $t = 0$, which is not integrable in $0$.)

Example — Damped sine absolutely convergent

For $t \ge 0$, $|\mathrm{e}^{-t} \sin(t)| \le \mathrm{e}^{-t}$, and $\int_0^{+\infty} \mathrm{e}^{-t}\,\mathrm{d}t = 1$ (reference of § 1.4). The integral is absolutely convergent.

Proposition — Endpoint reduction

Let $f \in \mathcal{C}_{pm}(]a, a+\varepsilon], \mathbb{K})$. Then $f$ is integrable in $a$ (meaning $\int_a^{a+\varepsilon} |f|$ converges as an improper integral) if and only if $t \mapsto f(a + t)$ is integrable in $0$ (on $]0, \varepsilon]$). The symmetric statement holds at $b$ with $t \mapsto f(b - t)$.

Proof

For every $x \in ]a, a+\varepsilon]$, translation invariance of the segment integral applied to the CM function $|f|$ on $[x, a+\varepsilon]$ (which follows from the continuous segment change of variable applied between consecutive jumps of $|f|$ + Chasles, recalled from Integration on a segment) gives $$ \int_x^{a+\varepsilon} |f(t)|\,\mathrm{d}t \;=\; \int_{x-a}^{\varepsilon} |f(a+u)|\,\mathrm{d}u. $$ Pass to the limit $x \to a^+$ (equivalently $x - a \to 0^+$): the two sides have the same nature and the same limit when convergent. Hence $\int_a^{a+\varepsilon} |f|$ converges if and only if $\int_0^{\varepsilon} |f(a + \cdot)|$ does.

Proposition — Reference $\int_a^b \mathrm{d}x/|x-a|^\alpha$ (corollary of § 1.4 + endpoint reduction)

For $\alpha \in \mathbb{R}$ and $a < b$ finite, $\int_a^b \mathrm{d}x/|x - a|^\alpha$ converges if and only if $\alpha < 1$. (Reduces to $\int_0^{b-a} \mathrm{d}t/t^\alpha$ by translation $t = x - a$.)

Method — Proving integrability by majoration

Bound $|f|$ by the general term of a known integrable reference (Riemann at $+\infty$ with $\alpha > 1$, Riemann at $0$ with $\alpha < 1$, exponential at $+\infty$, $|\ln t|$ at $0$). Then $\int_I |f|$ converges by the majoration rule, and $f$ is integrable. The integrability comparison theorem of § 2.3 generalises this method to $O$, $o$, $\sim$ comparisons.

Skills to practice

Proving integrability by majoration

II.3 Comparison theorem for integrability

This is the program § 8b comparison-for-integrability theorem: given a positive integrable reference $g$, an asymptotic comparison $f = O(g)$ / $f = o(g)$ / $f \sim g$ implies that $f$ is itself integrable. This is distinct from § 3, which starts from a comparison-on-the-integrand and transfers the asymptotic to the remainders or partial integrals. § 2.3 yields a conclusion about $f$'s integrability; § 3 yields an asymptotic equivalent for the integral itself.

Theorem — Comparison $O / o$ for integrability

Let $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$ and $g \in \mathcal{C}_{pm}([a, b[, \mathbb{R}^+)$, with $g$ integrable on $[a, b[$. If $f = O(g)$ (resp. $f = o(g)$) at $b^-$, then $f$ is integrable on $[a, b[$.

Proof

Hypothesis $f = O(g)$ at $b^-$ gives a rank $c \in [a, b[$ and $M \ge 0$ such that $|f(t)| \le M g(t)$ for $t \in [c, b[$. On $[c, b[$, the majoration rule (positive $|f|$ vs positive $M g$) gives integrability of $|f|$ on $[c, b[$. On the finite head $[a, c]$, $|f|$ is CM hence bounded, hence integrable. By Chasles, $|f|$ is integrable on $[a, b[$, so $f \in L^1([a, b[, \mathbb{K})$. The $o$ case is identical with $\varepsilon$ in place of $M$.

Theorem — Comparison $\sim$ for integrability

Let $f, g \in \mathcal{C}_{pm}([a, b[, \mathbb{R}^+)$ with $g$ eventually strictly positive near $b^-$. If $f \sim g$ at $b^-$, then $f$ is integrable on $[a, b[$ if and only if $g$ is.

Proof

$f \sim g$ with $g > 0$ eventually means $f/g \to 1$ at $b^-$, so from some rank $c$, $f/g \in [1/2, 2]$, i.e. $g/2 \le f \le 2g$. By two applications of the majoration rule, $f$ and $g$ are simultaneously integrable on $[c, b[$; the finite head $[a, c]$ adds a bounded CM contribution, which is automatically integrable. Conclude by Chasles.

Method — Choosing the comparison

At $+\infty$, compare with $1/t^\alpha$ for some $\alpha > 1$. Near $0$, compare with $1/t^\alpha$ for some $\alpha < 1$. Near a finite singularity $a$, perform the change of variable $t = x - a$ to reduce to the $0$ case.

Example — Gamma function: integrability

For $\alpha \in \mathbb{R}$, the Gamma integral $\int_0^{+\infty} t^{\alpha-1} \mathrm{e}^{-t}\,\mathrm{d}t$ is integrable in $0$ if and only if $\alpha > 0$ (compare with $t^{\alpha-1}$, $\int_0^1 t^{\alpha-1}\,\mathrm{d}t$ converges iff $-(\alpha-1) < 1$, i.e. $\alpha > 0$); and integrable in $+\infty$ for every $\alpha$ (by comparison with $\mathrm{e}^{-t/2}$: $t^{\alpha-1} \mathrm{e}^{-t} = o(\mathrm{e}^{-t/2})$ at $+\infty$ since $t^{\alpha-1} \mathrm{e}^{-t/2} \to 0$). So $\Gamma(\alpha) = \int_0^{+\infty} t^{\alpha-1} \mathrm{e}^{-t}\,\mathrm{d}t$ converges for every $\alpha > 0$.

Example — Logarithm over square root converges

The integrand has two potential singularities, at $0$ and at $1$.

Near $0$. $\sqrt{1 - t^2} \to 1$, so $\ln t/\sqrt{1 - t^2} \sim \ln t$. Take $g = -\ln t$ (the absolute value): $\int_0^1 (-\ln t)\,\mathrm{d}t = 1$ converges (reference of § 1.4), so $|\!\ln t / \sqrt{1 - t^2}|$ is integrable in $0$ by majoration.
Near $1$. $\sqrt{1 - t^2} = \sqrt{(1 - t)(1 + t)} \sim \sqrt{2(1 - t)}$ and $\ln t = \ln(1 + (t-1)) \sim -(1 - t)$, so the integrand behaves like $-(1 - t)/\sqrt{2(1 - t)} = -\sqrt{(1 - t)/2}$, which $\to 0$ as $t \to 1^-$. The singularity at $1$ is faussement impropre.

Both endpoints are settled, so $\int_0^1 \ln t / \sqrt{1 - t^2}\,\mathrm{d}t$ converges.

Skills to practice

Studying integrability by comparison

III Integration of comparison relations

Dictionary with chapter 13 § 2.3

This section is the integral mirror of Numerical and vector series § 2.3 (summation of comparison relations). The dictionary is exact: a series $\sum u_n$ has a partial sum $S_n$ (discrete index) and a remainder $R_n$ (defined when the series converges); an improper integral $\int_a^b f$ has a partial integral $\int_a^x f$ (continuous index $x \to b^-$) and a remainder $\int_x^b f$ (defined when the integral converges). The two summation theorems carry over verbatim: positive reference; transfer to remainders if integrable, to partial integrals if not. The only technical difference is the passage à la limite is continuous ($x \to b^-$) rather than discrete ($n \to +\infty$).

III.1 Integrable case

When the reference function $g$ is positive and integrable, an asymptotic comparison on the integrand transfers to the remainders. This is the mirror of the convergent-case summation of Numerical and vector series § 2.3.

Theorem — Integration of comparison relations -- integrable case

Let $g \in \mathcal{C}_{pm}([a, b[, \mathbb{R}^+)$ be integrable on $[a, b[$. At $b^-$:

($O$ and $o$ cases) For $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$: if $f = O(g)$ (resp. $f = o(g)$) at $b^-$, then $f$ is integrable on $[a, b[$, and the remainders satisfy $\int_x^b f = O(\int_x^b g)$ (resp. $o(\int_x^b g)$) as $x \to b^-$.
($\sim$ case) For $f \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ real-valued: if $f \sim g$ at $b^-$ (with $g > 0$ eventually, so $f > 0$ eventually too), then $f$ is integrable on $[a, b[$ and $\int_x^b f \sim \int_x^b g$.

Proof

Case $O$. From the comparison-for-integrability theorem of § 2.3, $f$ is integrable. Then for some $M \ge 0$ and rank $c \in [a, b[$, $|f(t)| \le M g(t)$ for $t \in [c, b[$. For $x \ge c$, the triangle inequality (Prop of § 2.2) gives $$ \left|\int_x^b f\right| \;\le\; \int_x^b |f| \;\le\; M \int_x^b g, $$ so $\int_x^b f = O(\int_x^b g)$.
Case $o$. Same argument with $\varepsilon$ in place of $M$: for any $\varepsilon > 0$, $|\int_x^b f| \le \varepsilon \int_x^b g$ from some rank.
Case $\sim$. $f$ is integrable by the § 2.3 $\sim$-comparison. Write $f = g + (f - g)$ with $f - g = o(g)$. By the $o$ case, $\int_x^b (f - g) = o(\int_x^b g)$, so $\int_x^b f - \int_x^b g = o(\int_x^b g)$, equivalently $\int_x^b f / \int_x^b g \to 1$, i.e. $\int_x^b f \sim \int_x^b g$.

Method — Summing an asymptotic comparison on integrable remainders

Pick a positive $g$ whose remainder asymptotic is known (a primitive of $g$ available), then transfer. Most often: $g(t) = 1/t^\alpha$ for $\alpha > 1$ at $+\infty$, with $\int_x^{+\infty} \mathrm{d}t/t^\alpha = 1/((\alpha-1) x^{\alpha-1})$ explicit.

Example — Riemann tail at order three

By direct primitive, $\int_x^{+\infty} \mathrm{d}t/t^3 = [-1/(2t^2)]_x^{+\infty} = 1/(2x^2)$, so $\int_x^{+\infty} \mathrm{d}t/t^3 \sim 1/(2x^2)$ as $x \to +\infty$ (equality, in fact). This is the building block for transferring $\sim$-comparisons on positive integrable references via the theorem above.

Example — Flagship -- the Gaussian asymptotic

Set $F(x) := \int_x^{+\infty} \mathrm{e}^{-t^2}\,\mathrm{d}t$. First, $F(x)$ converges absolutely: for $t \ge 1$, $t^2 \ge t$, so $\mathrm{e}^{-t^2} \le \mathrm{e}^{-t}$, and $\int_1^{+\infty} \mathrm{e}^{-t}\,\mathrm{d}t = 1/\mathrm{e}$ converges (\S 1.4 reference); on $[0, 1]$ the integrand is bounded continuous, hence integrable on any segment. So $\mathrm{e}^{-t^2}$ is integrable on $[0, +\infty[$ and $F(x)$ is well-defined for $x \ge 0$. Now apply one IPP with $u(t) = -1/(2t)$ and $v'(t) = -2t \mathrm{e}^{-t^2}$ (so $u'(t) = 1/(2t^2)$ and $v(t) = \mathrm{e}^{-t^2}$); then $u v' = \mathrm{e}^{-t^2}$. The standard IPP form $\int u v' = [uv] - \int u' v$ on $[x, +\infty[$ (for $x > 0$) gives $$ F(x) \;=\; \left[\frac{-\mathrm{e}^{-t^2}}{2t}\right]_x^{+\infty} \,-\, \int_x^{+\infty} \frac{\mathrm{e}^{-t^2}}{2t^2}\,\mathrm{d}t \;=\; \frac{\mathrm{e}^{-x^2}}{2x} \,-\, R(x), $$ where $R(x) := \int_x^{+\infty} \mathrm{e}^{-t^2}/(2t^2)\,\mathrm{d}t > 0$. Non-circular bound on $R$: for $t \ge x$, $1/t^2 \le 1/x^2$, so $$ 0 \;\le\; R(x) \;\le\; \frac{1}{2x^2} \int_x^{+\infty} \mathrm{e}^{-t^2}\,\mathrm{d}t \;=\; \frac{F(x)}{2x^2}. $$ Substituting back: $F(x)\bigl(1 + O(1/x^2)\bigr) = \mathrm{e}^{-x^2}/(2x)$, so $F(x) \sim \mathrm{e}^{-x^2}/(2x)$ as $x \to +\infty$. The non-circularity comes from the elementary bound $1/t^2 \le 1/x^2$, not from a self-referential integration of $o$.

Skills to practice

Integrating comparison relations -- integrable case

III.2 Non-integrable case

When the reference function $g$ is positive and not integrable on $[a, b[$, an asymptotic comparison transfers to the partial integrals (whose limit is $+\infty$). This is the mirror of the divergent-case summation of Numerical and vector series § 2.3.

Theorem — Integration of comparison relations -- non-integrable case

Let $g \in \mathcal{C}_{pm}([a, b[, \mathbb{R}^+)$ be not integrable on $[a, b[$ (so $\int_a^x g \to +\infty$). At $b^-$:

($O$ and $o$ cases) For $f \in \mathcal{C}_{pm}([a, b[, \mathbb{K})$: if $f = O(g)$ (resp. $f = o(g)$) at $b^-$, then the partial integrals satisfy $\int_a^x f = O(\int_a^x g)$ (resp. $o(\int_a^x g)$) as $x \to b^-$.
($\sim$ case) For $f \in \mathcal{C}_{pm}([a, b[, \mathbb{R})$ real-valued: if $f \sim g$ at $b^-$, then $f$ is also not integrable on $[a, b[$ and $\int_a^x f \sim \int_a^x g$.

Proof

Case $o$. For $\varepsilon > 0$, from some rank $c \in [a, b[$, $|f(t)| \le \varepsilon g(t)$ for $t \in [c, b[$. Split: $$ \left|\int_a^x f\right| \;\le\; \underbrace{\int_a^c |f|}_{\text{constant } C} \,+\, \int_c^x |f| \;\le\; C \,+\, \varepsilon \int_c^x g \;\le\; C \,+\, \varepsilon \int_a^x g. $$ Since $\int_a^x g \to +\infty$, eventually $C \le \varepsilon \int_a^x g$, so $|\int_a^x f| \le 2 \varepsilon \int_a^x g$. As $\varepsilon$ was arbitrary, $\int_a^x f = o(\int_a^x g)$.
Case $O$. Same splitting with a fixed constant $M$ replacing $\varepsilon$.
Case $\sim$. Write $f = g + (f - g)$ with $f - g = o(g)$. The $o$ case gives $\int_a^x (f - g) = o(\int_a^x g)$, so $\int_a^x f = \int_a^x g + o(\int_a^x g)$, i.e. $\int_a^x f \sim \int_a^x g$. Since $\int_a^x g \to +\infty$, so does $\int_a^x f$, so $f$ is not integrable on $[a, b[$ either.

Method — Equivalent of a partial integral

Pick a positive non-integrable $g$ with known partial-integral asymptotic. Typical references: $g(t) = 1/t$ at $+\infty$ (partial integral $\ln x$); $g(t) = 1/t^\alpha$ with $\alpha < 1$ at $+\infty$ (partial integral $x^{1-\alpha}/(1-\alpha)$); $g(t) = 1$ on $[a, +\infty[$ (partial integral $x - a$).

Example — The Bertrand integral

A primitive of $t \mapsto 1/(t \ln t)$ on $[2, +\infty[$ is $t \mapsto \ln(\ln t)$, so $$ \int_2^x \frac{\mathrm{d}t}{t \ln t} \;=\; \ln(\ln x) \,-\, \ln(\ln 2) \;\sim\; \ln(\ln x) \quad\text{as }x \to +\infty. $$ (Lower bound $2$ chosen to avoid the singularity of $1/(t \ln t)$ at $t = 1$.) Compare with the analogous series $\sum_{k \ge 2} 1/(k \ln k) \sim \ln(\ln n)$ of Numerical and vector series § 2.2 --- exact dictionary.

Example — Partial integral of $\ln t / t$

A primitive of $t \mapsto \ln t / t$ on $[1, +\infty[$ is $t \mapsto (\ln t)^2 / 2$, so $\int_1^x \ln t / t\,\mathrm{d}t = (\ln x)^2 / 2 \sim (\ln x)^2 / 2$ as $x \to +\infty$ (equality). This is a building block for $\sim$-transfer on non-integrable references.

Skills to practice

Integrating comparison relations -- non-integrable case