Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a$?

share|cite|improve this question
Perhaps because WolframAlpha is concerned that $a$ might be complex:^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. Dec 8 '10 at 0:03
up vote 24 down vote accepted

Yes it is true!


$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$

Make the substitution $\displaystyle x = \dfrac{a^2}{t}$

We get

$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$


$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$

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

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

share|cite|improve this answer
+1 Nice explanation ! – Ewin Feb 18 '15 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 $\pi/2$.

share|cite|improve this answer
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 $$ \int_{-\infty}^\infty \frac{1}{2}\cdot\frac{1}{1+t^2}\,dt=\frac{\pi}{2}. $$

share|cite|improve this answer
Fine. Up Vote $0$k. – Felix Marin Oct 17 '13 at 7:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.