- #1
ijustlost
- 22
- 0
I'm trying to find
[tex]
\int_{-\infty}^{\infty} \frac{exp(ax)}{cosh(x)} dx
[/tex]
where 0<a<1 and x is taken to be real. I'm doing this by contour integration using a contour with corners +- R, +- R + i(pi), and I'm getting an imaginary answer which is
[tex]\frac{2i\pi}{sin (a \pi)}[/tex].
I'm thinking this is a problem because my original integral was completely real. Can I just take the real part of my answer, and say the integral = 0 ? That doesn't seem to make any sense, I've drawn a graph of the function and it doesn't look like it's integral should be zero! I'm fairly sure my answer to the contour integral is correct!
[tex]
\int_{-\infty}^{\infty} \frac{exp(ax)}{cosh(x)} dx
[/tex]
where 0<a<1 and x is taken to be real. I'm doing this by contour integration using a contour with corners +- R, +- R + i(pi), and I'm getting an imaginary answer which is
[tex]\frac{2i\pi}{sin (a \pi)}[/tex].
I'm thinking this is a problem because my original integral was completely real. Can I just take the real part of my answer, and say the integral = 0 ? That doesn't seem to make any sense, I've drawn a graph of the function and it doesn't look like it's integral should be zero! I'm fairly sure my answer to the contour integral is correct!
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