Integral:$$\int_0^{\infty}\frac{\cos t}{(t^2+1)^2}dt$$
(Note: this might need methods of complex integration).

Question

Integral:$$\int_0^{\infty}\frac{\cos t}{(t^2+1)^2}dt$$
(Note: this might need methods of complex integration).

anonymous · Answer

I think first you have to convert cost somehow to a tan using sec^2+1=tan^2, then just use trig integral and do the bottom, something should be canceling out

anonymous · Answer

and Im sorry its sec^2-1=tan^2

anonymous · Answer

$$\int\limits_{0}^{n} \frac{ 1 }{ \sec(t) }*\frac{ 1 }{ (x^2+1)^2 }$$
then you just do the $$x=1*\tan \theta$$
$$dx = \sec^2(\theta) d \theta $$
$$x^2+1=\sec^2(\theta)$$
then you just substitute back everything

anonymous · Answer

@kirbykirby you know trig integral rule right?

anonymous · Answer

and that was 1/sec(x) not 1/sec(t) haha...

kirbykirby · Answer

But then it's something like $$ \int \frac{1}{\sec(\tan (\theta))}\frac{1}{\sec^2(\theta)}d\theta$$

zarkon · Answer

use residues

anonymous · Answer

I see... I didn't calculate it so I don't know that. But from something I find online http://memorize.com/half-angle-formulas it has something to do e^ix you can do some research on that