Improper Integrals 71

1. Core Concept Overview & Intuition

The definite integral studied earlier was built on the assumption that the interval of integration is finite and that the integrand is a bounded function on that interval. The theory of improper integrals removes one or both of these restrictions. An integral becomes improper in one of two ways: either the interval of integration is unbounded (one or both limits extend to infinity), or the integrand becomes unbounded at one or more points of the interval (a singularity lies inside or at an endpoint of the interval).

The central idea is to replace the offending limit or singularity by a parameter, integrate over a genuine (proper) interval, and then take a limit. Convergence of the improper integral is then defined as the existence of a finite limit of these proper integrals. This limiting procedure is the single unifying device behind every definition in this lesson.

Two distinct kinds of improper integral are treated. The improper integral of the first kind has an infinite interval of integration. The improper integral of the second kind has a finite interval but an unbounded integrand. The convergence tests, the notion of absolute and conditional convergence, and the Cauchy inequality apply to both kinds, while the Cauchy principal value is a device for assigning a meaningful value to certain divergent integrals by taking a symmetric limiting process.

Intuitively, an improper integral measures a total quantity (an accumulated area, in the geometric language of the earlier chapter on area under a curve) over an unbounded region. Convergence means that this total quantity is finite even though the region is unbounded; divergence means the accumulated quantity grows without bound or fails to settle at a single finite value.

2. Mathematical Formalization & Core Definitions

2.1 Objectives of the Lesson

> By the end of this lesson the learner should be able to:
> - state the definition of an improper integral of the first kind and of the second kind;
> - decide whether a given improper integral converges or diverges by direct evaluation of the defining limit;
> - distinguish absolute convergence from conditional convergence;
> - state and apply Cauchy's inequality for the convergence of an integral;
> - prove and use the result that absolute convergence of an integral implies convergence;
> - compute the Cauchy principal value of an improper integral with a singularity.

2.2 Definition of Improper Integrals

> Definition (Improper integral of the first kind). Let $f$ be a function defined on the interval $[a,\infty)$ and integrable on every finite subinterval $[a,b]$ with $b>a$. The improper integral of $f$ over $[a,\infty)$ is defined by

$$
\int_{a}^{\infty} f(x)\,dx=\lim_{b\to\infty}\int_{a}^{b} f(x)\,dx.
$$

> If this limit exists and is finite, the improper integral is said to converge; otherwise it is said to diverge.

> Definition (Improper integral of the first kind, lower limit). For a function $f$ defined on $(-\infty,b]$ and integrable on every $[a,b]$,

$$
\int_{-\infty}^{b} f(x)\,dx=\lim_{a\to-\infty}\int_{a}^{b} f(x)\,dx.
$$

> Definition (Improper integral of the first kind, doubly infinite interval). For a function $f$ defined on $(-\infty,\infty)$ and integrable on every finite subinterval, the integral is defined by splitting at any real number $c$:

$$
\int_{-\infty}^{\infty} f(x)\,dx=\int_{-\infty}^{c} f(x)\,dx+\int_{c}^{\infty} f(x)\,dx,
$$

> and the integral converges only when both integrals on the right converge.

> Definition (Improper integral of the second kind). Let $f$ be defined on $[a,b)$ and unbounded in every neighbourhood of $b$, but integrable on every $[a,t]$ with $a\le t<b$. Then

$$
\int_{a}^{b} f(x)\,dx=\lim_{t\to b^{-}}\int_{a}^{t} f(x)\,dx,
$$

> and the integral converges if this limit exists and is finite, and diverges otherwise. An analogous definition holds when the singularity is at the lower endpoint $a$, using $\lim_{t\to a^{+}}\int_{t}^{b} f(x)\,dx$. If the singularity lies at an interior point $c$ with $a<c<b$, the integral is defined as

$$
\int_{a}^{b} f(x)\,dx=\int_{a}^{c} f(x)\,dx+\int_{c}^{b} f(x)\,dx,
$$

> and converges only when both parts converge.

In both kinds the improper integral is therefore a limit of proper integrals. The integrand $f$ is called the integrand of the improper integral, and the variable $x$ is the variable of integration.

2.3 Convergence and Divergence of the Improper Integrals of the First Kind

The behaviour of the improper integral of the first kind is governed by the rate at which the integrand decays as the variable tends to infinity. The model case is the power function.

> Result (Power test of the first kind). For $a>0$ and real $p$,

$$
\int_{a}^{\infty}\frac{1}{x^{p}}\,dx
$$

> converges when $p>1$ and diverges when $p\le 1$.

The proof rests on direct evaluation of the defining limit. For $p\neq 1$ and $b>a$,

$$
\int_{a}^{b}\frac{1}{x^{p}}\,dx=\frac{b^{1-p}-a^{1-p}}{1-p}.
$$

If $p>1$ then $1-p<0$, so $b^{1-p}\to 0$ as $b\to\infty$ and the limit is $\dfrac{a^{1-p}}{p-1}$, a finite number. If $p<1$ then $1-p>0$, so $b^{1-p}\to\infty$ and the integral diverges. For $p=1$,

$$
\int_{a}^{b}\frac{1}{x}\,dx=\ln b-\ln a,
$$

which tends to infinity with $b$, so the integral diverges. The same reasoning applied to the exponential function gives convergence of $\int_{a}^{\infty} e^{-kx}\,dx$ for every $k>0$, since the defining limit of $\dfrac{e^{-ka}-e^{-kb}}{k}$ is finite.

Divergence of an improper integral of the first kind may occur in two ways: the defining limit may be infinite, or it may fail to exist at all. For example, $\int_{0}^{\infty}\sin x\,dx$ diverges because $\int_{0}^{b}\sin x\,dx=1-\cos b$ oscillates without approaching a limit as $b\to\infty$.

2.4 Definition of Absolute and Conditional Convergence

> Definition (Absolute convergence). The improper integral $\int_{a}^{\infty} f(x)\,dx$ is said to converge absolutely if the integral of the absolute value,

$$
\int_{a}^{\infty}\lvert f(x)\rvert\,dx,
$$

> converges.

> Definition (Conditional convergence). The improper integral $\int_{a}^{\infty} f(x)\,dx$ is said to converge conditionally if it converges but does not converge absolutely, that is, $\int_{a}^{\infty} f(x)\,dx$ converges while $\int_{a}^{\infty}\lvert f(x)\rvert\,dx$ diverges.

The same definitions apply verbatim to improper integrals of the second kind, with the singularity replacing the infinite limit. Absolute convergence is a stronger requirement than convergence: it forbids the cancellation of positive and negative contributions from being the sole reason the integral settles at a finite value.

2.5 Cauchy's Inequality for Convergence of an Integral

> Theorem (Cauchy's inequality). Let $f$ and $g$ be functions for which the integrals $\int_{a}^{\infty} f^{2}(x)\,dx$ and $\int_{a}^{\infty} g^{2}(x)\,dx$ converge. Then the integral $\int_{a}^{\infty} f(x)g(x)\,dx$ converges absolutely, and

$$
\left\lvert\int_{a}^{\infty} f(x)g(x)\,dx\right\rvert\le\left(\int_{a}^{\infty} f^{2}(x)\,dx\right)^{1/2}\left(\int_{a}^{\infty} g^{2}(x)\,dx\right)^{1/2}.
$$

The inequality is the integral analogue of the Cauchy–Schwarz inequality for finite sums. It is used as a convergence test: if the two square-integrals converge, then the product integral converges absolutely, and the displayed bound controls its magnitude. The inequality is applied in practice by choosing $f$ and $g$ so that the product $fg$ is the given integrand and each of $f^{2}$ and $g^{2}$ is a power or exponential whose convergence is already settled by the power test.

2.6 Absolute Convergence of an Integral Implies Convergence

> Theorem (Absolute convergence implies convergence). If $\int_{a}^{\infty}\lvert f(x)\rvert\,dx$ converges, then $\int_{a}^{\infty} f(x)\,dx$ converges, and

$$
\left\lvert\int_{a}^{\infty} f(x)\,dx\right\rvert\le\int_{a}^{\infty}\lvert f(x)\rvert\,dx.
$$

The proof uses the comparison of the positive and negative parts of $f$. Write $f=f^{+}-f^{-}$, where $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$ are non-negative. Since $0\le f^{+}\le\lvert f\rvert$ and $0\le f^{-}\le\lvert f\rvert$, the convergence of $\int_{a}^{\infty}\lvert f\rvert\,dx$ forces the convergence of $\int_{a}^{\infty} f^{+}\,dx$ and $\int_{a}^{\infty} f^{-}\,dx$ by comparison. Hence $\int_{a}^{\infty} f\,dx=\int_{a}^{\infty} f^{+}\,dx-\int_{a}^{\infty} f^{-}\,dx$ converges, and the displayed bound follows from $\lvert f\rvert=f^{+}+f^{-}$. The converse of the theorem is false: conditional convergence shows that convergence need not imply absolute convergence.

2.7 Convergence and Divergence of the Improper Integrals of the Second Kind

The improper integral of the second kind is governed by the rate at which the integrand blows up at the singularity. The model case is again a power function.

> Result (Power test of the second kind). For $a<b$ and real $p$,

$$
\int_{a}^{b}\frac{1}{(b-x)^{p}}\,dx
$$

> converges when $p<1$ and diverges when $p\ge 1$. Similarly,

$$
\int_{a}^{b}\frac{1}{(x-a)^{p}}\,dx
$$

> converges when $p<1$ and diverges when $p\ge 1$.

The proof evaluates the defining limit. For $p\neq 1$ and $a\le t<b$,

$$
\int_{a}^{t}\frac{1}{(b-x)^{p}}\,dx=\frac{(b-a)^{1-p}-(b-t)^{1-p}}{1-p}.
$$

If $p<1$ then $1-p>0$, so $(b-t)^{1-p}\to 0$ as $t\to b^{-}$ and the limit is $\dfrac{(b-a)^{1-p}}{1-p}$, finite. If $p>1$ then $1-p<0$, so $(b-t)^{1-p}\to\infty$ and the integral diverges. For $p=1$,

$$
\int_{a}^{t}\frac{1}{b-x}\,dx=\ln(b-a)-\ln(b-t),
$$

which tends to infinity as $t\to b^{-}$, so the integral diverges. Note the reversal of the exponent threshold relative to the first kind: for the second kind convergence requires $p<1$, whereas for the first kind it requires $p>1$.

2.8 The Cauchy Principal Value of Improper Integrals with Singularities

When an improper integral diverges only because of a symmetric cancellation failure, a symmetric limiting procedure may still assign it a finite value.

> Definition (Cauchy principal value, singularity at an interior point). Let $f$ have a singularity at $c$ with $a<c<b$. The Cauchy principal value of $\int_{a}^{b} f(x)\,dx$ is

$$
\mathrm{PV}\int_{a}^{b} f(x)\,dx=\lim_{\varepsilon\to 0^{+}}\left(\int_{a}^{c-\varepsilon} f(x)\,dx+\int_{c+\varepsilon}^{b} f(x)\,dx\right),
$$

> provided this limit exists.

> Definition (Cauchy principal value, doubly infinite interval). For a function $f$ defined on $(-\infty,\infty)$,

$$
\mathrm{PV}\int_{-\infty}^{\infty} f(x)\,dx=\lim_{R\to\infty}\int_{-R}^{R} f(x)\,dx,
$$

> provided this limit exists.

The principal value differs from the improper integral itself. The improper integral requires the two one-sided limits to exist independently, whereas the principal value ties them together through a single symmetric parameter. Consequently a principal value may exist even when the improper integral diverges. For example, $\int_{-1}^{1}\dfrac{1}{x}\,dx$ diverges as an improper integral of the second kind, because each of $\int_{0}^{1}\dfrac{1}{x}\,dx$ and $\int_{-1}^{0}\dfrac{1}{x}\,dx$ diverges. But the symmetric limit gives

$$
\mathrm{PV}\int_{-1}^{1}\frac{1}{x}\,dx=\lim_{\varepsilon\to 0^{+}}\left(\int_{-1}^{-\varepsilon}\frac{1}{x}\,dx+\int_{\varepsilon}^{1}\frac{1}{x}\,dx\right)=\lim_{\varepsilon\to 0^{+}}\left(-\ln\varepsilon+\ln\varepsilon\right)=0.
$$

The two divergent logarithmic contributions cancel exactly under the symmetric limiting process, so the principal value is $0$. The principal value is therefore a regularisation device, not a substitute for convergence: whenever the improper integral converges, its principal value exists and equals it, but the converse fails.