# 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 $a$?

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Perhaps because WolframAlpha is concerned that $a$ might be complex: http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%28%28x^2-i^2%29^2%2‌​Bx^2%29+dx+for+x+%3D+0..infinity – Rahul Dec 7 '10 at 20:23
You can certainly evaluate it via contour integration. – Robin Chapman Dec 7 '10 at 20:34
In general, Wolfram Alpha (and Mathematica) always assume that variables are complex, unless told otherwise (e.g. with Assuming[]). – J. M. is back. Dec 8 '10 at 0:03

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

$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$

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+1 Nice explanation ! – Ewin Feb 18 at 14:23

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 $\pi/2$.

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For general complex $a$ there are still 2 poles in upper halfplane, but not necessarily $x_1$ and $x_2$, so the answer may be different (and dependent on $a$). – Max Dec 8 '10 at 0:53

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 $$\int_{-\infty}^\infty \frac{1}{2}\cdot\frac{1}{1+t^2}\,dt=\frac{\pi}{2}.$$

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Fine. Up Vote $0$k. – Felix Marin Oct 17 '13 at 7:51