Improper integrals and trig substitution

[zje]

Homework Statement

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:

<br /> \int \limits_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\textrm{d}x<br />

a is a constant

Homework Equations

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

The Attempt at a Solution

I make the substitution

x = a \textrm{tan} \theta

therefore

<br /> \textrm{d}x = a\textrm{sec}^2\theta\textrm{d}\theta<br />

giving me

<br /> \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}}<br />

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...

Thanks for your help!

 

[zje]

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...

 

[dextercioby]

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.

 

[spamiam]

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 into get expressions in x and then use the original limits. Or you can find the new limits for theta by solving
<br /> \infty = x = a \tan\theta<br />
and
<br /> -\infty = x = a \tan\theta<br />
for theta. (Hint: To do the second method, think about where cosine is 0, and whether sine is positive or negative at this point.)

 

[zje]

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