# Show $\int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ -\pi < a < \pi$

I want to show that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ -\pi < a < \pi$

So I let $f \displaystyle (z) = e^{ibx} \frac{\sinh az}{\sinh \pi z}$ and integrated under the rectangular contour with vertices at $R, R+ i, R + i$, and $-R$, and and an indentation at $z=i$ to avoid the simple pole.

I end up with $\displaystyle \text{PV} \left( \int_{-\infty}^{\infty} f(x) \ dx + e^{-b} \cos a \int_{-\infty}^{\infty} f(t) \ dt + i e^{-b} \sin a \int_{-\infty}^{\infty} e^{ibt} \frac{\cosh at}{\sinh \pi t} \ dt \right) = e^{-b} \sin a$

And if I equate real parts, $\displaystyle (1+ e^{-b} \cos a) \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx - e^{-b}\sin a \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt = e^{-b} \sin a$

Did I use wrong contour or the wrong function? Or is there a relationship between $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx$ and $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt$?

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– Sam Mar 24 '12 at 20:58
Thanks.  It's essentially the same problem.  But how did he evaluate that infinite series? – Random Variable Mar 24 '12 at 21:15
$(-1)^n e^{-n\pi\omega}\sin(\pi \kappa n)$ is the imaginary part of $\exp(-n\pi \omega + i \pi \kappa n + i\pi n) = e^{-n\pi (\omega +i (\kappa + 1))}$, which is a geometric series. – Sam Mar 24 '12 at 22:26
I should have realized that. – Random Variable Mar 24 '12 at 23:06
But I think the contour should be a rectangle with vertices at $N + \frac{1}{2}, (N+ \frac{1}{2})+i(N+\frac{1}{2}),-(N+\frac{1}{2}) + i(N+ \frac{1}{2})$, and $-N-\frac{1}{2}$ where $N$ is an integer – Random Variable Mar 24 '12 at 23:13

## 1 Answer

I found another approach which is much longer but easier to justify.

Let $\displaystyle f(z) = \frac{e^{(a+ib)z}}{\sinh \pi z}$

Notice that $\displaystyle f(z) = \frac{(\cosh az + \sinh az)e^{ibz}}{\sinh \pi z}$

Now integrate along the rectangle with vertices at $-R, R, R+i,$ and $-R+i$, and indentations at $z=0$ and $z=i$.

After a lot of work, you will end up not only showing that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx = \frac{\sin a}{\cos a + \cosh b}$ , but also that $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx = \frac{\sinh b}{\cos a + \cosh b}$.

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