What is improper integral? — Mathematics

Definition 9.1 University Mathematics — Year 2 · Chapter 9 — Integration

Let $f$ be piecewise continuous on $\intco{a}{b}$ ( $b \in \R$ or $+\infty$ ). The integral converges when $\lim_{x \to b^-} \int_a^x f$ exists; one then writes $\int_a^b f$ for the limit. (Similarly on $\intoc{a}{b}$ , and on $\intoo{a}{b}$ by splitting at an interior point — the choice does not matter, by Chasles.) The integral converges absolutely when $\int_a^b \abs f$ converges; absolute convergence implies convergence, by the Cauchy criterion:

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

and completeness of $\R$ (the primitive has the Cauchy property). In detail: let $F(x) = \int_a^x f$ and $G(x) = \int_a^x \abs f$ . If $\int^b\abs f$ converges, $G$ has a limit at $b^-$ , so for every $\varepsilon > 0$ there is $c < b$ with $G(y) - G(x) \leq \varepsilon$ whenever $c \leq x \leq y < b$ ; the display transfers this Cauchy property to $F$ . For any sequence $x_n \to b^-$ the values $F(x_n)$ then form a Cauchy sequence of reals, convergent by completeness, and interlacing two such sequences shows the limit is the same for all of them: $F$ has a limit at $b^-$ .

Examples

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

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

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

(b) $\displaystyle\int_0^\infty \frac{\ln t}{1 + t^2}\,\dd t$ : trouble at both ends, so split at $1$ . Near $0$ : $\abs{\ln t}$ integrable as in (a); near $\infty$ : $\frac{\ln t}{1+t^2} = o(t^{-3/2})$ : absolutely convergent. The substitution $t = \frac1u$ maps $\intoo{0}{1}$ onto $\intoo{1}{\infty}$ and

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

the two halves cancel, and the integral is $0$ . Closing insight: symmetry under $t \mapsto \frac1t$ is worth a page of computation — the same trick already powered Exercise 9.3.

Example 9.4 (One value, three integrals)

Study $I = \displaystyle\int_0^{\infty} \frac{1 - \cos t}{t^2}\,\dd t$ . At $0$ : $1 - \cos t \sim \frac{t^2}2$ , so the integrand extends continuously by the value $\frac12$ — no singularity at all. At $\infty$ : $0 \leq \frac{1 - \cos t}{t^2} \leq \frac{2}{t^2}$ : absolute convergence (Theorem 9.2). Value: integrate by parts on $\intcc{\varepsilon}{M}$ with $u = 1 - \cos t$ , $v' = t^{-2}$ :

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

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

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

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

Example 9.5 (A semi-convergent integral)

$\displaystyle\int_1^{\infty} \frac{\sin t}{t}\,\dd t$ converges: integrate by parts,

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

where the bracket has a limit and the last integral converges absolutely ( $\abs{\cos t}/t^2 \leq t^{-2}$ ). But not absolutely: from $\abs{\sin t} \geq \sin^2 t$ ,

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

where the last integral converges by the same integration by parts as above (with $\sin 2t$ in the bracket): a divergent piece minus a convergent one diverges. So $\int_1^\infty\frac{\sin t}{t}\dd t$ converges without converging absolutely — the integral analogue of the alternating series, with integration by parts playing the role of the alternating test.

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