
Though contour integral
I'm trying to crack this problem but I can't seem to get it right.
http://i54.photobucket.com/albums/g9...f/complex1.jpg
It's the third exercise from from chapter 2 of "Complex Analysis" by Stein & Shakarchi.
I've drawn the sector (contour C) here (sorry for the crappy MSpaint picture):
http://i54.photobucket.com/albums/g9...f/complex2.jpg
As there are no poles within the sector I set $\displaystyle \oint_{C}f(z)dz = 0$, so that leaves
$\displaystyle 0 = \int_{0}^{A}e^{Ax}dx + \int_{0}^{w}iAe^{A^2e^{i\theta}}e^{i\theta}d\theta + \int_{0}^{A}e^{iw}e^{Ate^{iw}}dt$
Now, I assume one of these three components goes to zero as A goes to infinity and indeed the second one (with $\displaystyle \theta$ as variable) does. Then I'm stuck because what I find $\displaystyle \int_{0}^{A = x = \infty}e^{Ax}dx = 1/A$ doesn't lead me to anything remotely like the answer Wolfram's online calculator gave me if I use integration by parts
http://i54.photobucket.com/albums/g9...f/complex3.jpg
Can anyone help me with this problem, what am I doing wrong?

I'm not familiar with complex analysis, so this probably won't be of help, but both can be done by using byparts (twice).

I know, but I don't think I'd get away with that if this problem was in my exam.

Let $\displaystyle f(z) = e^{Az }~~,~~ A = \sqrt{a^2 + b^2 } $
Consider the path $\displaystyle C$ as you have drawn , since $\displaystyle f(z)$ is analytic everywhere , we have $\displaystyle \int_C e^{Az}~dz = 0 $
$\displaystyle \int_{C_1} e^{Az} ~dz + \int_{C_2} e^{Az} ~dz + \int_{C_3} e^{Az} ~dz = 0 $
$\displaystyle C_1 $ is the straight line starting from $\displaystyle O$ to $\displaystyle R ~( R + 0i ) $
$\displaystyle C_2 $ is the arc $\displaystyle R \to R e^{i \omega} ~~ \omega = \angle( a+ib) $
$\displaystyle C_3 $ is a straight line $\displaystyle R e^{i \omega} \to O $
Then we have $\displaystyle \int_{C_1} e^{Az} ~dz = \int_0^R e^{Ax}~dx = \frac{1}{A} $
$\displaystyle 0 < \int_{C_2} e^{Az}~dz < \int_{C_2} e^{Az}~dz = \int_{C_2} e^{AR} ~dz \to 0 ~~ \text{as}~ R \to \infty $
$\displaystyle \int_{C_2} e^{Az} ~dz = 0 $
$\displaystyle \int_{C_3} e^{Az} ~dz $
Sub $\displaystyle z = x e^{i \omega} $
we have
$\displaystyle \int_0^R e^{Ax e^{i \omega} } ~ e^{i \omega} ~dx $
but $\displaystyle A e^{i \omega} = a + bi ~ \implies e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} } $
so
$\displaystyle \int_{C_3} e^{Az} ~dz = \frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{ax  ibx} ~dx$
By joining the three integrals we obtain
$\displaystyle \frac{1}{A} + 0 \frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{ax  ibx} ~dx = 0$
$\displaystyle \frac{1}{a+bi} = \int_0^{\infty} e^{ax} ( \cos(bx)  i \sin(bx) )~dx = \frac{abi}{a^2+b^2 } $
By comparing the real part and the Im part , we have
$\displaystyle \int_0^{\infty} e^{ax} \cos(bx) ~dx = \frac{a}{a^2+b^2} $
$\displaystyle \int_0^{\infty} e^{ax} \sin(bx) ~dx = \frac{b}{a^2+b^2} $

Thank you very much, the $\displaystyle A e^{i \omega} = a + bi ~ \implies e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} }$ part was the missing link. Every contour integral needs its own set of tricks it seems, very annoying, but at least I can have some peace of mind now.