Theory

Theory 1

Improper integrals are those for which either a bound or the integrand itself become infinite somewhere on the interval of integration.

Examples:

$$(a){\int }_1^{\infty }\frac{1}{x^2}dx,(b){\int }_0^2\frac{1}{x}dx,(c){\int }_{-1}^{+1}\frac{1}{x^2}dx$$

(a) the upper bound is $\infty$ (b) the integrand goes to $\infty$ as $x\to 0^{+}$ (c) the integrand is $\infty$ at the point $0\in [-\mathrm{1,1}]$

The limit interpretation of (a) is this:

$${\int }_1^{\infty }\frac{1}{x^2}dx=\underset{R\to \infty }{\lim }{\int }_1^R\frac{1}{x^2}dx$$

The limit interpretation of (b) is this:

$${\int }_0^2\frac{1}{x}dx=\underset{R\to 0^{+}}{\lim }{\int }_R^2\frac{1}{x}dx$$

The limit interpretation of (c) is this:

$$\begin{array}{c}{\int }_{-1}^{+1}\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^{+1}\frac{1}{x^2}dx\\ \\ =\underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x^2}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x^2}dx\end{array}$$

These limits are evaluated using the usual methods.

An improper integral is said to be convergent or divergent according to whether it may be assigned a finite value through the appropriate limit interpretation.

For example, $(a)$ converges while $(b)$ diverges.

Theory 2

Two tools allow us to determine convergence of a large variety of integrals. They are the comparison test and the $p$-integral cases.

Comparison test - integrals

The comparison test says:

• When an improper integral converges, every smaller integral converges.
• When an improper integral diverges, every bigger integral diverges.

Here, smaller and bigger are comparisons of the integrand at all values (accounting properly for signs), and the bounds are assumed to be the same.

For example, ${\int }_2^{\infty }\frac{dx}{x^3}$ converges, and $x^4>x^3$ implies $\frac{1}{x^4}<\frac{1}{x^3}$ (when $x>1$), therefore the comparison test implies that ${\int }_2^{\infty }\frac{dx}{x^4}$ converges.

$p$-integral cases

Assume $p>0$ and $a>0$. We have:

$$\begin{array}{cccc}p>1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{converges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{diverges}\\ & & & \\ p<1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{converges}\\ & & & \\ p=1:& {\int }_a^{\infty }\frac{dx}{x}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x}\text{diverges}\end{array}$$

Proving the $p$-integral cases

It is easy to prove the convergence / divergence of each $p$-integral case using the limit interpretation and the power rule for integrals. (Or for $p=1$, using $\int \frac{1}{x}dx=\ln x+C$.)

Additional improper integral types

The improper integral ${\int }_{-\infty }^{a}f(x)dx$ also has a limit interpretation:

$${\int }_{-\infty }^{a}f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx$$

The double improper integral ${\int }_{-\infty }^{\infty }f(x)dx$ has this limit interpretation:

$${\int }_{-\infty }^{\infty }f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx+\underset{R\to \infty }{\lim }{\int }_a^{R}f(x)dx$$

Where $a$ is any finite number. This double integral does not exist if either limit does not exist for any value of $a$.

Double improper is not simultaneous!

Watch out! This may happen:

$${\int }_{-\infty }^{\infty }f(x)dx\ne \underset{R\to \infty }{\lim }{\int }_{-R}^{R}f(x)dx$$

This simultaneous limit might exist only because of internal cancellation in a case where the separate individual limits do not exist! We do not say ‘convergent’ in these cases!