TITLE: Is this definite integral really independent of a parameter? How can it be shown?
QUESTION [18 upvotes]: I want to find a nice simple expression for the definite integral
$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$
Now, I can numerically compute this integral, and it seems to converge to $\pi/2$ for all real values of $a$. Is this integral actually always equal to $\pi/2$? How can I show this?
Also, why does Wolfram Alpha give me something that appears to depend on $a$? Is there a good reason it doesn't eliminate $a$?

REPLY [5 votes]: Aryabhata's solution is nice. 
The method of residue is standard in complex function theory.
Here it is a simple elementary derivation.
We may assume that $a\ge 0$.
$$
\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}=\int_0^\infty \frac{1}{1+\left( x-\frac{a^2}{x} \right)^2}\,dx.
$$
If we had 
$$
\int_0^\infty \frac{1}{1+t^2}\,dt
$$
then we could calculate it easily.
This motivates the substitution
$$
x-\frac{a^2}{x}=:t\, \qquad(1).
$$
Here
$$
D_x\left( x-\frac{a^2}{x} \right)=1+\frac{a^2}{x^2}\gt 0, \qquad (x\gt 0).
$$
From $(1)$ we obtain
$$
x=\frac{t}{2}+\frac{1}{2}\sqrt{t^2+4 a^2}
$$
because $x\gt0$.
From this
$$
dx=\left( \frac{1}{2}+\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+4a^2}} \right)\,dt.
$$
Substituting back into the integral we get
$$
\int_0^\infty \frac{1}{1+\left( x-\frac{a^2}{x} \right)^2}\,dx=
\int_{-\infty}^\infty \left(\frac{1}{2}+\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+4a^2}}\right)\frac{1}{1+t^2}    \,dt
$$
Here the second integrand is an odd function so the result is
$$
\int_{-\infty}^\infty \frac{1}{2}\cdot\frac{1}{1+t^2}\,dt=\frac{\pi}{2}.
$$