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

Question

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 $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt$?

Answer 1

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 $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx = \frac{\sinh b}{\cos a + \cosh b}$.