# Though contour integral

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• Jul 5th 2010, 10:06 AM
Gulli
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 MS-paint picture):
http://i54.photobucket.com/albums/g9...f/complex2.jpg

As there are no poles within the sector I set $\oint_{C}f(z)dz = 0$, so that leaves
$0 = \int_{0}^{A}e^{-Ax}dx + \int_{0}^{w}iAe^{-A^2e^{i\theta}}e^{i\theta}d\theta + \int_{0}^{A}e^{iw}e^{-Ate^{iw}}dt$

Now, I assume one of these three components goes to zero as A goes to infinity and indeed the second one (with $\theta$ as variable) does. Then I'm stuck because what I find $\int_{0}^{A = x = \infty}e^{-Ax}dx = 1/A$ 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?
• Jul 5th 2010, 10:18 AM
TheCoffeeMachine
I'm not familiar with complex analysis, so this probably won't be of help, but both can be done by using by-parts (twice).
• Jul 5th 2010, 11:42 AM
Gulli
I know, but I don't think I'd get away with that if this problem was in my exam.
• Jul 5th 2010, 08:02 PM
simplependulum
Let $f(z) = e^{-Az }~~,~~ A = \sqrt{a^2 + b^2 }$

Consider the path $C$ as you have drawn , since $f(z)$ is analytic everywhere , we have $\int_C e^{-Az}~dz = 0$

$\int_{C_1} e^{-Az} ~dz + \int_{C_2} e^{-Az} ~dz + \int_{C_3} e^{-Az} ~dz = 0$

$C_1$ is the straight line starting from $O$ to $R ~( R + 0i )$

$C_2$ is the arc $R \to R e^{i \omega} ~~ \omega = \angle( a+ib)$

$C_3$ is a straight line $R e^{i \omega} \to O$

Then we have $\int_{C_1} e^{-Az} ~dz = \int_0^R e^{-Ax}~dx = \frac{1}{A}$

$0 < |\int_{C_2} e^{-Az}~dz| < \int_{C_2} e^{-A|z|}~|dz| = \int_{C_2} e^{-AR} ~|dz| \to 0 ~~ \text{as}~ R \to \infty$

$\int_{C_2} e^{-Az} ~dz = 0$

$\int_{C_3} e^{-Az} ~dz$

Sub $z = x e^{i \omega}$

we have

$-\int_0^R e^{-Ax e^{i \omega} } ~ e^{i \omega} ~dx$

but $A e^{i \omega} = a + bi ~ \implies e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} }$

so
$\int_{C_3} e^{-Az} ~dz = -\frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{-ax - ibx} ~dx$

By joining the three integrals we obtain

$\frac{1}{A} + 0 -\frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{-ax - ibx} ~dx = 0$

$\frac{1}{a+bi} = \int_0^{\infty} e^{-ax} ( \cos(bx) - i \sin(bx) )~dx = \frac{a-bi}{a^2+b^2 }$

By comparing the real part and the Im part , we have

$\int_0^{\infty} e^{-ax} \cos(bx) ~dx = \frac{a}{a^2+b^2}$

$\int_0^{\infty} e^{-ax} \sin(bx) ~dx = \frac{b}{a^2+b^2}$
• Jul 6th 2010, 08:54 AM
Gulli
Thank you very much, the $A e^{i \omega} = a + bi ~ \implies e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} }$ 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.