1. ## Complex integration

integration from 0->oo cosx-1/x^2dx=-pi/2

show that ^. the x^2 at the bottom throws me off because your points equal to zero?? Im not sure how to start?

2. Originally Posted by durham2
integration from 0->oo cosx-1/x^2dx=-pi/2

show that ^. the x^2 at the bottom throws me off because your points equal to zero?? Im not sure how to start?

What did you mean to write: cos(-1)x/x^2 or (cosx -1)/x^2 or cos x - (1/x^2) (this last is what most people would usually understand, but it is trivial so I think it must be the first or second option)...or something else??

Tonio

3. It is the second one!

4. So it's:

$\displaystyle \int_0^{\infty} \frac{\cos(x)-1}{x^2}$

Usually when you have sines and cosines like that, consider using $\displaystyle e^{iz}=\cos(z)+i\sin(z)$ then in your case, take the real part. So how about trying:

$\displaystyle \mathop\oint\limits_{W_u} \frac{e^{iz}}{z^2}dz$

With $\displaystyle W_h$ being the upper half-washer with an indentation around the pole at the origin. Note, the pole is simple, so we can immediately calculate the integral around the indentation as it's radius goes to zero. It's $\displaystyle -\pi i r_0$ where $\displaystyle r_0$ is the residue at the origin. Add the other two legs from minus infinity to plus infinity and note that the integrand is even so take one half of that.