# Improper integrals and trig substitution

1. ### zje

16
1. The problem statement, all variables and given/known data
It's been a couple of years since I've done real math, so I'm kinda stuck on this one. This is actually part of a physics problem, not a math problem - but I'm stuck on the calculus part. I'm trying to solve this guy:

$\int \limits_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\textrm{d}x$

a is a constant

2. Relevant equations

$\textrm{tan}^2 \theta + 1 = \textrm{sec}^2 \theta$

3. The attempt at a solution
I make the substitution

$x = a \textrm{tan} \theta$

therefore

$\textrm{d}x = a\textrm{sec}^2\theta\textrm{d}\theta$

giving me

$\int{\frac{a^2 \textrm{tan}^2 \theta a\textrm{sec}^2 \theta \textrm{d} \theta}{(a^2 \textrm{tan}^2 \theta + a^2)^2}}$

and eventually I get it to boil down to (using the aforementioned tangent identity and canceling terms)
$\frac{1}{a} \int \textrm{tan}^2 \theta \textrm{d} \theta$
I thought I was supposed to change the limits to
$\pm\frac{\pi}{2}$
, but when I solve the above simplified integral I get
$\textrm{tan}\theta - \theta$
which is not convergent

My problem is taking the limit for the tangent at $\pm\frac{\pi}{2}$
I'm probably screwing up with the limits of integration. What exactly am I supposed to do with a trig substitution and the limits when dealing with an improper integral? I was following an old calculus book of mine, but this doesn't seem exactly right...

Last edited: Feb 9, 2011
2. ### zje

16
Just realized that it shouldn't end at
$\textrm{tan}^2\theta$
but
$\frac{1}{a} \int \frac{\textrm{tan}^2\theta}{\textrm{sec}^2\theta}\textrm{d}\theta = \frac{1}{a} \int \textrm{sin}^2 \theta \textrm{d}\theta$
I'm still unsure of what exactly to do with the limits...

3. ### dextercioby

12,314
I'm getting

$$\int_{-\infty}^{\infty} \frac{x^2}{\left(x^{2}+a^{2}\right)^{2}} dx = \int_{0}^{\infty} x \, \frac{2x}{\left(x^{2}+a^{2}\right)^{2}} dx = ...$$

and now you can do part integration.

4. ### spamiam

366
To solve your resulting integral $\int sin^2\theta \, d\theta$, use the half-angle formula on $sin^2 \theta$. As for the limits of integration, there are two possibilities. Once you've integrated, you can substitute back in to get expressions in x and then use the original limits. Or you can find the new limits for theta by solving
$$\infty = x = a \tan\theta$$
and
$$-\infty = x = a \tan\theta$$
for theta. (Hint: To do the second method, think about where cosine is 0, and whether sine is positive or negative at this point.)

5. ### zje

16
Thanks all for your help, I think I got it!