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

I tried integrating $f \displaystyle (z) = e^{ibz} \, \frac{\sinh az}{\sinh \pi z} $ around an indented rectangular contour with vertices at $\pm R, \pm R+ i$.

I ended up with the equation $$ (1+ e^{-b} \cos a) \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \, \cos bx \, dx - e^{-b}\sin a \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \, \sin bx \ dx = e^{-b} \sin a . $$

But I don't see how to extract the value of the integral from this equation.

Did I integrate the wrong function?

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Possible duplicate of (math.stackexchange.com/questions/77172/…). 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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I indeed integrated the wrong function.

Integrating $ \displaystyle f(z) = \frac{e^{(a+ib)z}}{\sinh \pi z}$ around the same contour, we get

$$ \begin{align} \text{PV} \int_{-\infty}^{\infty} f(x) \, dx + e^{-b}e^{ia} \ \text{PV}\int_{-\infty}^{\infty} f(x) \ dx &= \pi i \, \text{Res}[f(z),0] + \pi i \ \text{Res}[f(z),i]\\ &= i \, \left(1-e^{-b}e^{ia} \right) . \end{align} $$

Therefore,

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

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

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

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

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