Submitted to HEJ Manuscript no.: ANM-010201-A

Improper integrals

The notion of integral as defined above is rather restrictive, many extensions and generalizations have been given. Here two simple extensions will be presented, one to unbounded functions and another one to infinite intervals. Instead of stating general definitions and theorems we shall look at some specific examples.

Consider the function

$$f\ :\ x\rightarrow \frac {1}{\sqrt{x}}$$

on $[0,1]$. The function $f$ is not well-defined in $x=0$. Let us put $f(0)=0$ (or any other real value). Now look at

$>$ Int(1/sqrt(x),x=y..1) = int(1/sqrt(x),x=y..1);

$${\displaystyle \int _{y}^{1}} {\displaystyle \frac {1}{\sqrt{x}} } \,dx=2 - 2\,\sqrt{y}$$

where $y$ is a positive number less than $1$. Obviously this expression has $2$ as the limit when $y\rightarrow 0$. Define

$$\int _{0}^{1}\frac {1}{\sqrt{x}}\,dx = \lim _{y\rightarrow 0}\,\int _{y}^{1}\frac {1}{\sqrt{x}}\,dx.$$

Then we have

$$\int _{0}^{1}\frac {1}{\sqrt{x}}\,dx= 2.$$

The integral on the left is called an improper integral. Now let us see the extension to infinite intervals. Look at the following example:

$>$ Int(1/x^2,x=1..y)=int(1/x^2,x=1..y);

$${\displaystyle \int _{1}^{y}} {\displaystyle \frac {1}{x^{2}}} \, dx= - {\displaystyle \frac {1}{y}} + 1$$

The right-hand member has limit $1$ for $y\rightarrow \infty$. Defining

$$\int _{1}^{\infty }\frac {1}{x^{2}}\,dx = \lim _{y\rightarrow \infty }\,\int _{1}^{y}\frac {1}{x^{2}}\,dx$$

we get

$$\int _{1}^{\infty }\frac {1}{x^{2}}\,dx= 1.$$

The left-hand side is again called an improper integral. Next we have an example of a doubly improper integral:

$>$ Int(exp(-x)*(2*x-1)/sqrt(x),x=0..infinity);

$${\displaystyle \int _{0}^{\infty }} {\displaystyle \frac {e^{( - x)}\,(2\,x - 1)}{\sqrt{x}}} \,dx$$

The interval is infinite and the integrand is unbounded near $x=0$. Now for $0<y<z$ look at

$>$ Int(exp(-x)*(2*x-1)/sqrt(x),x=y..z)=value(%);

$${\displaystyle \int _{y}^{z}} {\displaystyle \frac {e^{( - x... ...}}} \,dx= - 2\,\sqrt{z}\,e^{( - z)} + 2\,\sqrt{ y}\,e^{( - y)}$$

When $y\rightarrow 0$ and $z\rightarrow \infty$ (independently), the right-hand side tends to $0$. Hence

$$\int _{0}^{\infty }\frac {e^{( - x)}\,(2\,x - 1)}{\sqrt{x}}\,dx = 0.$$

There are many interesting improper integrals, e.g. the Dirichlet integral

$>$ Int(sin(x)/x,x=0..infinity);

$${\displaystyle \int _{0}^{\infty }} {\displaystyle \frac {{\rm sin}(x)}{x}} \,dx$$

$>$ value(%);

$${\displaystyle \frac {1}{2}} \,\pi.$$

Maple doesn't compute this integral but knows it. We shall skip the proof.
Another famous one is the gamma-function, defined by

$$\Gamma(x) = \int _{0}^{\infty }t^{(x - 1)}\,e^{( - t)}\,dt$$

for all $x\geq 1$.

Exercise 5.1.

Prove that

• $\Gamma(1)=1$,
• $\Gamma(x+1)= x \Gamma(x)$,
• $\Gamma(n)=(n-1)!$ when $n$ is a positive integer.

Hint: integration by parts.

Last but not least, some good advice: use your brains and think twice before computing. The integral

$$\int _{-\pi}^{\pi} \frac{\sin t}{1+\sin ^8 t}dt$$

seems to be hard to evaluate but there is also no need. You can almost immediately conclude that the integrand is odd and therefore the integral must be equal to zero.

Submitted to HEJ Manuscript no.: ANM-010201-A