- #1
eschiesser
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Homework Statement
I need to solve this integral for a>0:
[itex]\int _0^{\infty }\frac{\text{Sin}[x]}{x}\frac{1}{x^2+a^2}dx[/itex]
The Attempt at a Solution
Using wolfram mathematica, I get that this integral is:
[itex]\frac{\pi -e^{-x} \pi }{2x^2}=\frac{\pi (1-\text{Cosh}[a]+\text{Sinh}[a])}{2 a^2}[/itex]
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
[itex]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}[/itex]
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!