# Thread: Though contour integral

1. ## Though contour integral

I'm trying to crack this problem but I can't seem to get it right.

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):

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

Can anyone help me with this problem, what am I doing wrong?

2. 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).

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

4. 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 $\displaystyle \int_{C_1} e^{-Az} ~dz = \int_0^R e^{-Ax}~dx = \frac{1}{A}$

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

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

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

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

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

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

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

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

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

5. 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.