Is this definite integral really independent of a parameter? How can it be shown? 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 $a$?
 A: 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
$$
\int_{-\infty}^\infty \frac{1}{2}\cdot\frac{1}{1+t^2}\,dt=\frac{\pi}{2}.
$$
A: Yes it is true! 
Let
$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
Make the substitution $\displaystyle x = \dfrac{a^2}{t}$
We get
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$
i.e.
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$
Therefore
$$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$
$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$
Making the substitution $\displaystyle t = x - \dfrac{a^2}{x}$
Gives us
$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$
A: With apologies to Robin Chapman.
The integrand is an even function of $x$, so we can integrate from $-\infty$ to $\infty$ and take half. The integrand tends to $1/z^2$ for large $z$ and as the length of a large arc is $\pi z$ the contribution of the arc tends to zero. So we just need to integrate over the upper half plane. The residues are solutions to $0=(x^2-a^2)^2+x^2= (x^2-i x -a^2)(x^2+i x-a^2)$. If $x_1$ and $x_2$ are solutions to the first quadratic, then at $x_1$ the second polynomial is equal to $2i x_1$, and the residue at $x_1$ is $x_1^2/2ix_1(x_2-x_1)=x_1/2i(x_2-x_1)$. The sum of residues at $x_1$ and $x_2$ is therefore $1/2i$. Now we just note that the two poles in the upper half plane are indeed the solutions $x_1$ and $x_2$ (which are  $i (\pm\sqrt{-a^2+1/4} + 1/2)$)). Hence the contour integral is $\pi$, and the original integral is $\pi/2$.
