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Q: If $|a|< 1$ and $b>0$, show that $$\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx = \frac{\pi \sin (\pi a)}{2 (\cos (\pi a)+\cosh (\pi b))}$$

I need to evaluate the above integral by method of contour. I tried to use this contour on this question but at $2\pi i$, $\sinh(ax)$ changes to $\sinh(ax+2a\pi i)$ and I have difficulty taking out $\sinh(ax)$. Please give hints on which contour to use.Thanks in advance!!

ADDED::
Considering $-R \to R \to R + \pi i \to -R + \pi i \to -R$ with a bump on $0$ and $\pi i$ to avoid singularity.

$$(1 + e^{(a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(a+ib)\pi i}) \hspace{1 cm}(1)$$ $$(1 + e^{(-a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(-a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(-a+ib)\pi i}) \hspace{1 cm}(2)$$ With a bit of algebra, we get \begin{align*} \int_{-\infty}^{\infty}\frac{e^{ax}-e^{-ax}}{\sinh x}e^{ibx}dx &= 2\pi i \left( \frac{1}{1 + e^{(a+bi)\pi i}} - \frac{1}{1 + e^{(-a+bi)\pi i}} \right)\\ &= 2 \pi \frac{\sin (a\pi)}{\cosh (b\pi) + \sin(a\pi)} \end{align*} From which we get the desired result.

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2 Answers 2

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Use parity to extend the domain of integration from $-\infty$ to $\infty$. Shift the contour of integration down by $\frac{i\pi}{2}$ to avoid pole $x=0$. Then compute separately four integrals (actually, it suffices to compute one of them and then to change parameters accordingly) $$\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(\pm a\pm ib)x}dx}{\sinh x}$$ using the rectangle $$-R-\frac{i\pi}{2}\rightarrow R-\frac{i\pi}{2}\rightarrow R+\frac{i\pi}{2} \rightarrow -R+\frac{i\pi}{2} \rightarrow -R-\frac{i\pi}{2}$$ with $R\rightarrow\infty$.


For example, let us compute \begin{align} \int_{\text{rectangle}}\frac{e^{(a+ib)z}dz}{\sinh z}\substack{R\rightarrow\infty\\=}\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}- \int_{-\infty+\frac{i\pi}{2}}^{\infty+\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}=\\ =\left(1+e^{\pi i (a+ib)}\right)\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z} \end{align} On the other hand, the integral over rectangle is equal to $2\pi i$ (the only pole inside is $z=0$ and the residue is $1$). Therefore, $$\int_{-\infty-\frac{i\pi}{2}}^{\infty-\frac{i\pi}{2}}\frac{e^{(a+ib)z}dz}{\sinh z}=\frac{2\pi i}{1+e^{\pi i (a+ib)}},$$ and the initial integral evaluates to $$\frac{2\pi i}{8}\left(\frac{1}{1+e^{\pi i (a+ib)}}-\frac{1}{1+e^{\pi i (-a+ib)}}+\frac{1}{1+e^{\pi i (a-ib)}}-\frac{1}{1+e^{\pi i (-a-ib)}}\right)=\frac{\pi\sin\pi a}{2\left(\cos\pi a+\cosh\pi b\right)}.$$

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  • $\begingroup$ sorry .. could you elaborate? what happens from $R \to R + 2\pi i$ as $R \to \infty$ does it go to zero? $\endgroup$ Commented May 18, 2013 at 20:24
  • $\begingroup$ yes, of course. On the segment $z=R+is$, $s\in[0,2\pi]$ the integrand becomes $$\frac{e^{(a+ib)z}}{\sinh z}=2\frac{e^{(a+ib)(R+is)}}{e^{R+is}-e^{-R-is}}=O\left(e^{-(1-a)R}\right)\qquad \text{as}\;R\rightarrow\infty.$$ One also has similar estimate on the left vertical segment . $\endgroup$ Commented May 18, 2013 at 20:33
  • $\begingroup$ I have written a complete solution (it turns out to be easier to use rectangle of height $i\pi$ instead of $2i\pi$. Hope now it's clear. $\endgroup$ Commented May 18, 2013 at 21:33
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Consider

$$f(z) = \frac{\sinh(ax)e^{ibx}}{\sinh(x)}$$

so that the function we want integrated is $\Re f$. Let us find

$$I = \int_{-\infty}^\infty f(z)\,dz$$

which is twice the integral in question. Let $C$ be this integral, indented to avoid the pole at $z=0$: Contour

The integral around the large arc disappears as $R\to\infty$, and as $\epsilon\to 0$, the integral around the small arc becomes:

$$\pi i \operatorname*{Res}_{z=0} f(z) = 0$$

Then we have, seeing that $f$ has poles in $C$ at $z_n = \pi i n$ for $n \ge 1$.

$$\operatorname*{Res}_{z=\pi in} f(z) = \lim_{z\to \pi i n}\frac{\sinh(ax)e^{ibx}}{\cosh(x)} = i(-1)^n e^{-b\pi n}\sin(\pi a n)$$

Then summing:

$$ I=\oint_C f(z)\, dz =\\ \Re\, 2\pi i \sum_{n=1}^\infty \operatorname*{Res}_{z=\pi i n} f(z) = \\ -2\pi\Im \sum_{n=1}^\infty (-1)^n e^{-b\pi n+\pi i a n} = \\ -2 \pi \Im \frac{1}{1+e^{-b\pi +\pi i a }} = \\ \frac{\pi\sin(a\pi)}{\cos(a\pi)+\cosh(b\pi)} $$

Divide by two and the answer is obtained.

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