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Improper integrals and trig substitution

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zje
#1
Feb8-11, 11:55 PM
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:

[itex]
\int \limits_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\textrm{d}x
[/itex]

a is a constant

2. Relevant equations

[itex]\textrm{tan}^2 \theta + 1 = \textrm{sec}^2 \theta[/itex]


3. The attempt at a solution
I make the substitution

[itex] x = a \textrm{tan} \theta[/itex]

therefore

[itex]
\textrm{d}x = a\textrm{sec}^2\theta\textrm{d}\theta
[/itex]

giving me

[itex]
\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}}
[/itex]

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

My problem is taking the limit for the tangent at [itex]\pm\frac{\pi}{2}[/itex]
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!
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zje
#2
Feb9-11, 12:19 AM
P: 16
Just realized that it shouldn't end at
[itex]\textrm{tan}^2\theta[/itex]
but
[itex]\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[/itex]
I'm still unsure of what exactly to do with the limits...
dextercioby
#3
Feb9-11, 03:51 AM
Sci Advisor
HW Helper
P: 11,927
I'm getting

[tex] \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 = ... [/tex]

and now you can do part integration.

spamiam
#4
Feb9-11, 08:37 AM
P: 366
Improper integrals and trig substitution

To solve your resulting integral [itex] \int sin^2\theta \, d\theta [/itex], use the half-angle formula on [itex] sin^2 \theta [/itex]. 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
[tex]
\infty = x = a \tan\theta
[/tex]
and
[tex]
-\infty = x = a \tan\theta
[/tex]
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
#5
Feb9-11, 05:46 PM
P: 16
Thanks all for your help, I think I got it!


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