Improper integrals and trig substitution

by zje
Tags: improper, integrals, substitution, trig
 P: 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... Thanks for your help!
 P: 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...
 Sci Advisor HW Helper P: 11,863 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.
P: 366

Improper integrals and trig substitution

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.)
 P: 16 Thanks all for your help, I think I got it!

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