Complex Analysis (Residual Theorem).

• Dec 4th 2008, 06:31 PM
illusion419
Complex Analysis (Residual Theorem).

This is what I did... I'm not sure if I'm in the right direction about this problem.

I am french so excuse me if my translation in english isn't accurate.

I used the residual theorem.

The integral of http://www.cramster.com/Answer-Board...0000003333.gif is 0. (Cauchy's Theorem).

Poles of order 1 at http://www.cramster.com/Answer-Board...4093750666.gif

Only the pole at z = ia is in the semi-circle drawn above.

Residual at z = ia :

What does cos(ia) represent ? Does that even exist?

If this is totally wrong, then how do I solve this problem?
• Dec 4th 2008, 11:25 PM
Moo
Hello,

$\cos(ia)=\frac{e^{i(ia)}+e^{-i(ia)}}{2}=\frac{e^{a}+e^{-a}}{2}=\cosh(a)$
• Dec 5th 2008, 05:08 AM
shawsend
Quote:

Originally Posted by illusion419

The integral of http://www.cramster.com/Answer-Board...0000003333.gif is 0. (Cauchy's Theorem).

That's not Cauchy's Theorem. You'd need to show that as the radius goes to infinity, the integral over the half-circle arc goes to zero. But it's easier if you use:

$\mathop\oint\limits_{H_u}\frac{e^{iz}}{z^2+a^2}dz= 2\pi i\mathop\text{Res}\limits_{z=ia}\left\{\frac{e^{iz }}{z^2+a^2}\right\}=f+gi$

where $H_u$ is the upper half circle contour. I think it's not to hard to show:

$\lim_{R\to\infty}\mathop\int\limits_C \frac{e^{iz}}{z^2+a^2}dz\to 0$

where $C$ is the half-arc contour and on the straight-line segment along the real axis $R$ we have:

$\mathop\int\limits_{R} \frac{e^{iz}}{z^2+a^2}dz=\int_{-\infty}^{\infty} \frac{cos(x)}{x^2+a^2}dx+i\int_{-\infty}^{\infty} \frac{sin(x)}{x^2+a^2}dx=f+gi$ (since the part over C is zero).

Then:

$\int_{-\infty}^{\infty} \frac{cos(x)}{x^2+a^2}dx=\textbf{Re}\left[2\pi i\mathop\text{Res}\limits_{z=ia}\left\{\frac{e^{iz }}{z^2+a^2}\right\}\right]$

 I made a silly mistake on this initially but corrected it here. Hope that didn't cause problems for you Illusion.