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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^{ibz} \frac{\sinh az}{\sinh \pi z} $ and integrated around a rectangular contour with vertices at $R, R+ i, -R + i$, and $-R$, and with an indentation around $z=i$ to avoid the simple pole.

I eventually end up with $$ (1+ e^{-b} \cos a) \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx - e^{-b}\sin a \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt = e^{-b} \sin a . $$

Did I use the wrong contour or perhaps the wrong function?

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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Possible duplicate of (…). I think the contour used by Sasha in his answer yields a very nice solution. It's certainly much cleaner! ;) – 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
up vote 7 down vote accepted

I have a simple way to calculate your old question. Note $$ \frac{\sinh(ax)}{\sinh(\pi x)} \cos(bx) = \frac{\sinh(ax)}{\sinh(\pi x)}\cosh(ibx)=\frac{\sinh((a+bi)x)+\sinh((a-bi)x)}{\sinh(\pi x)} $$ and $$ \int_0^\infty\frac{\sinh(ax)}{\sinh(bx)}dx=\frac{\pi}{2b}\tan\frac{a\pi}{2b}. $$ So \begin{eqnarray} I&=&\int_0^\infty\frac{\sinh(ax)}{\sinh(\pi x)} \cos(bx)dx\\ &=&\int_0^\infty\frac{\sinh((a+bi)x)+\sin((a-bi)x)}{\sinh(\pi x)}dx\\ &=&\frac{1}{2}\tan\frac{a+bi}{2}+\frac{1}{2}\tan\frac{a-bi}{2}\\ &=&\frac{\sin a}{\cos a + \cosh b}. \end{eqnarray}

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Nice answer. I only vaguely remember asking this question. – Random Variable Aug 9 '14 at 1:46

Another approach is to consider $ \displaystyle f(z) = \frac{e^{(a+ib)z}}{\sinh \pi z}$ and integrate around a rectangle with vertices at $-R, R, R+i,$ and $-R+i$, and half-circle indentations around the simple poles at $z=0$ and $z=i$.

Then since $\displaystyle \int f(z) \ dz$ vanishes on the left and right sides of the rectangle as $R \to \infty$, $$ \begin{align} \text{PV} \int_{-\infty}^{\infty} f(x) \ dx + e^{-b}e^{ia} \ \text{PV}\int_{-\infty}^{\infty} f(t) \ dt &= \pi i \ \text{Res}[f(z),0] + \pi i \ \text{Res}[f(z),i]\\ &= i \left(1-e^{-b}e^{ia} \right) . \end{align} $$


$$ \text{PV} \int_{-\infty}^{\infty} f(x) \ dx = i \ \frac{1-e^{-b} e^{ia}}{1+e^{-b}e^{ia}} = \frac{\sin a}{\cos a + \cosh b} + i \frac{\sinh b}{\cos a + \cosh b} .$$

But $$ \begin{align} \text{PV} \int_{-\infty}^{\infty} f(x) \ dx &= \text{PV} \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \cos bx \ dx + i \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx \ dx + \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx \\ &+ i \int_{-\infty}^{\infty} \frac{\sinh ax}{ \sinh \pi x} \sin bx \ dx \\ &= 0 + i \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx \ dx + \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx + 0 \\ &= \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx + i \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx \ dx . \end{align}$$

So equating the real parts on both sides of the equation, $$ \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx= \frac{\sin a}{\cos a + \cosh b}.$$

And equating the imaginary parts on both sides of the equation, $$ \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx \ dx = \frac{\sinh b}{\cos a + \cosh b}.$$

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