Definite integrals: solving with residue theory and contour integration

[eschiesser]

Homework Statement

I need to solve this integral for a>0:

\int _0^{\infty }\frac{\text{Sin}[x]}{x}\frac{1}{x^2+a^2}dx

The Attempt at a Solution

Using wolfram mathematica, I get that this integral is:

\frac{\pi -e^{-x} \pi }{2x^2}=\frac{\pi (1-\text{Cosh}[a]+\text{Sinh}[a])}{2 a^2}

I cannot get to this result. I have tried to split the sine into exponentials, and use the upper/lower infinite semi-circle contours, which results in

I =\frac{\pi }{2} \text{Res}\left[\frac{ e^{-i x}}{x}\frac{1}{x^2+a^2},\left\{x,a e^{i \frac{3\pi }{2}}\right\}\right]-\frac{\pi }{2} \text{Res}\left[\frac{ e^{i x}}{x}\frac{1}{x^2+a^2},\left\{x,a e^{i \frac{\pi }{2}}\right\}\right]=\frac{\pi \text{Sinh}[a]}{2 a^2}

I also tried to add a branch point and solve with that method but then I get the same answer...

Does the "removable singularity" at z=0 have something to do with this? Please steer me in the right direction!

 

[clamtrox]

If you split sine into exponentials and then split the integrals, then each integral has a pole at x=0 as well.

 

[eschiesser]

But then, isn't my contour running over the pole in each integral?

 

[jackmell]

But then, isn't my contour running over the pole in each integral?

Tell you what, how about you forget that problem for now and just try and solve:

\oint \frac{e^{iz}}{z}\frac{1}{z^2+a^2} dz

where the contour is the upper half-disc with an indentation around the pole at the origin. Take the limit as the radius of the indentation goes to zero and let the radius of the large arc go to infinity. Then you have:

\int_{-\infty}^{\infty} f(x)dx+\lim_{\rho\to 0}\int_{\pi}^0 f(z)dz+\lim_{R\to\infty}\mathop\int_{\text{half-arc}} f(z)dz=2\pi i r

Won't that work? See if it does. I don't know for sure.