# Evaluation of a real integral using Contour Integration

• May 20th 2011, 10:10 AM
garunas
Evaluation of a real integral using Contour Integration
I'm having difficulty in using contour Integration to evaluate the integral:

$\int_{0}^\infty \frac{cos(x)}{x^2 + a^2} dx$ where a>0.

Firstly I found that the function being integrated is an even function so this is equivalent to finding:

$\frac{1}{2} \int_{-\infty}^\infty \frac{cos(x)}{x^2 + a^2} dx$

Also $cos(x) = Re( exp(ix))$ where Re dentoes the Real part of exp(ix) function. The integral in question is now:
$\frac{1}{2}Re\int_{-\infty}^\infty \frac{exp(ix)}{x^2 + a^2} dx$
so the complex function in question is:
$f(z)=\frac{exp(iz)}{z^2 + a^2}$ which is holomorphic everywhere in the complex plane except for the points ai, and -ai. That is what I've found so far.

What contour do I have to use? Also how would you integrate it because I'm seeing there will be much difficulty integrating $\frac{exp(iz)}{z^2 + a^2}$ with the z being replaced by the suitable parameterization.
Thanks
• May 20th 2011, 10:11 AM
garunas
Also apologies if this thread is in the wrong section. I wasn't sure whether to put this under calculus or something else...
• May 20th 2011, 10:21 AM
Mondreus
Make a semicircle that includes the real axis. You can choose either one that encloses z=ia or z=-ia. Use the parametrization $z=Re^{it}$ for the semicircle (z=ia) and show what happens when R goes to infinity.
• May 20th 2011, 10:30 AM
garunas
Does it have to be a semi cirlce? Why not a full circle?
• May 20th 2011, 12:55 PM
TheEmptySet
Quote:

Originally Posted by garunas
Does it have to be a semi cirlce? Why not a full circle?

Because you want to use the residue theorem to find the value of the of the integral on

$C_1$ The part on the x axis (this is what you are looking for) and

$C_2$ the part on the circle above (or below) the x axis. You will want to show that this is equal to zero then

$\oint_{C_1+C_2}f(z)dz=\int_{C_1}f(z)dz+\int_{C_2}f (z)dx$

If the 2nd integral goes to zero and the first integral is over the x-axis you have found the value you are looking for.

See this example on wiki

Residue theorem - Wikipedia, the free encyclopedia
• May 20th 2011, 03:59 PM
garunas
That really helped a lot! Thank you :)