# integral Cauchy theorem

• Sep 30th 2007, 02:27 PM
integral Cauchy theorem
Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B
• Sep 30th 2007, 07:45 PM
topsquark
Quote:

Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B

Pretend this is an integration over a complex variable z:
$\int_{-\infty}^{+\infty}\frac{1}{z^2-2z+5}dz$.

Now, the Cauchy theorem says that if you integrate a complex integrand (counterclock-wise) over a closed loop that the integral is $2\pi i$ times the sum of the residues.

So what closed loop to use? Well, two leap to mind:
1) Start at $-\infty$ on the real axis and go to $\infty$, then close the loop by traveling a half-circle (of radius $\infty$ ) upward and to the left, finishing at $-\infty$ on the real axis. (Technically we go from -R to R on the real axis, and take the limit of the integrals as R goes to $\infty$.)

2) Same idea as the first, but close the loop by doing a half-circle downward and to the left.

In each case the ideal result is that the integral over the half-circle ends up being 0. This is the case here. (This is Jordan's Lemma.)

Which contour you choose depends on the integral and location of the poles. In this case you have one pole in the upper half-plane and one in the lower half-plane. I see no reason to choose one contour over the other here. So pick one and do it. (Or if you want some practice, pick both and do them to see that you get the same answer.)

-Dan
• Oct 1st 2007, 06:41 AM
Thank you. I will try that now!!
• Oct 6th 2007, 05:33 PM
ThePerfectHacker
Quote:

Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B

Define $f(z) = \frac{1}{z^2-2z+5} = -4i \left( \frac{1}{z-1-2i} - \frac{1}{z-1-2i} \right)$

Integrate over a large semicircle,
$-4i \int_{-R}^R \frac{1}{z-1-2i} - \frac{1}{z-1-2i} dz +\mbox{ error }$ $= 2 \pi i(-4i)[f'(1+2i) - f'(1-2i) ]$

By Cauchy integral theorem.

The error term goes to zero as $R\to \infty$.
• Oct 7th 2007, 06:16 AM
Thank you ThePerfectHacker!
• Oct 7th 2007, 10:22 AM
ThePerfectHacker
Quote:

Thank you ThePerfectHacker!

Would you like me do explain what I mean by "error" and why it goes to zero if you do not understand? (I was just sick yesterday and it was hard for me to respond fully).
• Oct 7th 2007, 03:22 PM
Yes please explain me this point,

I have not noticed the "error" before and I was about to ask you this question. I was suscpecting a term with a e^(-R* "something") that lead to zero when R-> inf. but I couls not find it!
• Oct 7th 2007, 07:03 PM
galactus
I am just delving into CA and don't want to appear anal, but I believe PH means:

$\frac{-i}{4}\left[\frac{1}{z-(1+2i)}-\frac{1}{z-(1-2i)}\right]$

This can be broken into two integrals, giving:

Which gives $\frac{-i}{4}(2{\pi}i)=\frac{\pi}{2}$