How can I calculate this integral $\int_{-\infty}^{\infty} d k \frac{\sinh a k}{\sinh (\pi k) \sinh (\frac{\pi k}{2})}e^{ikx}$

I have the following integral as: $$\int_{-\infty}^{\infty} d k \frac{\sinh a k}{\sinh (\pi k) \sinh (\frac{\pi k}{2})}e^{ikx}$$ where $$0\leq a \leq \frac{\pi}{2}$$.
Can any one help me calculataing this integral? Many thanks!

• Are you familiar with contour integration? Apr 8, 2021 at 16:54
• Could $a \in (-3 \pi/2,3 \pi/2)$? Apr 8, 2021 at 18:55
• Not exactly, but the approach is the same (integration in the complex plane) math.stackexchange.com/questions/395822/… or math.stackexchange.com/questions/61605/… Apr 8, 2021 at 20:53
• @RonGordon Yes it could be. Apr 8, 2021 at 21:32
• @MarkViola just a little bit, any hint will be appreciated. Apr 11, 2021 at 9:49

$$J=\int_{-\infty}^{\infty} \frac{\sinh a k}{\sinh (\pi k) \sinh (\frac{\pi k}{2})}e^{ikx}dk=\int_{-\infty}^{\infty} \frac{\sinh a k}{2\sinh^2 (\frac{\pi k}{2}) \cosh (\frac{\pi k}{2})}e^{ikx}dk=\int_{-\infty}^{\infty} \frac{\sinh 2a k}{\sinh^2 (\pi k) \cosh (\pi k)}e^{2ikx}dk$$

For the convenience we will evaluate the integral

$$J=\int_{-\infty}^{\infty} \frac{\sinh a k}{\sinh^2 (\pi k) \cosh (\pi k)}e^{ikx}dk=\frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{ikx}}{\sinh^2 (\pi k) \cosh (\pi k)}\Bigl(e^{ak}-e^{-ak}\Bigr)dk=\frac{1}{2}\bigr(I(a)-I(-a)\bigl)$$

We will consider $$I(a)$$ and $$I(-a)$$ separately. This is not rigorous, but we know that at $$a \in (-3 \pi/2,3 \pi/2)$$ integral is convergent, and we don't have singularity at $$x\to0$$: due to the symmetry only $$i\sin(kx)$$ survive from $$e^{ikx}$$, what gives additional power of $$x$$ at $$x\to0$$ and makes $$x=0$$ a removable singular point. $$I(a)=\int_{-\infty}^{\infty} \frac{e^{ikx}}{\sinh^2 (\pi k) \cosh (\pi k)}e^{ak}dk$$

We choose the following contour in the complex plane:

To close the contour we also add two small half-circles ($$C_1$$ and $$C_2$$ above the point $$x=0$$ and below $$x=i$$ (we go counter clockwise in both cases) and vertical lines $$1$$ and $$2$$ at $$R\to\pm\infty$$

Integral along the closed contour looks $$\oint =I(a)+\int_{C_1}+\int_1+\,\,e^{ia-x}I(a)+\int_{C_2}+\int_2=2\pi i \operatorname{Res}_{x=\frac{i}{2}}\biggl(\frac{e^{ikx}}{\sinh^2 (\pi k) \cosh (\pi k)}e^{ak}\biggr)$$

It can be shown that $$\int_1$$ and $$\int_2\, \to0$$ at $$R\to\infty$$. We have a single pole of first order inside the contour at $$x=\frac{i}{2}$$, so $$I(a)(1+e^{ia-x})+\int_{C_1}+\int_{C_2}=-2\exp{(\frac{ia-x}{2})}$$

$$\int_{C_1}=\lim_{r\to0}\int_{-\pi}^0\frac{\exp{(are^{i\phi})}\exp{(ix re^{i\phi})}}{\bigl(\pi re^{i\phi}+\frac{(\pi re^{i\phi})^3}{3!}+...\bigr)^2}ire^{i\phi}d\phi=\lim_{r\to0}\int_{-\pi}^0\frac{1+re^{i\phi}(a+ix)+ O(r^2)}{\bigl(\pi re^{i\phi}+\frac{(\pi re^{i\phi})^3}{3!}+...\bigr)^2}ire^{i\phi}d\phi$$

Integral contains the divergent term, but it does not depend on $$a$$, so it will be canceled when we evaluate $$I(a)-I(-a)$$ - as it should be. Keeping only converging terms we get: $$\int_{C_1}=-\frac{i}{\pi} (a+ix)$$

In the same fashion we get $$\int_{C_2}=\frac{i}{\pi} (a+ix)e^{ia-x}$$

Taking all together $$I(a)=\frac{i}{\pi}\,(a+ix)\frac{1-e^{ia-x}}{1+e^{ia-x}}-2\frac{e^{\frac{ia-x}{2}}}{1+e^{ia-x}}$$ $$I(a)-I(-a)=\frac{2ia}{\pi}\frac{\sinh x}{\cosh x+\cos a}+\frac{2 ix}{\pi}\frac{\sin a}{\cosh x+\cos a}-4i\frac{\sin\frac{a}{2}\sinh \frac{x}{2}}{\cosh x+\cos a}$$

Taking one half and switching to the initial $$a$$ and $$x$$ ($$a\to 2a$$ and $$x\to 2x$$) $$J=\frac{2i}{\pi}\,\frac{a\sinh 2x+x\sin 2a-\pi\sin a\sinh x}{\cosh 2x+\cos 2a}$$

• Just corrected several typos (lost $\frac{1}{\pi^2}$ in the first term - $\sinh^2\pi x=(\pi x)^2+...$ at $x\to0\,$) Apr 11, 2021 at 17:10
• First of all thank you very much @Svyatoslav for your detailed answer. I guess a minus sign is lost on the way $2\pi i Res_{x=\frac{i}{2}}=2\pi i (\frac{i}{\pi}e^{\frac{1}{2}(i a-x)})=-2e^{\frac{1}{2}(i a-x)}$. I think also that there is an extra pre-factor of 1/2 in the second definition of I(a) (just before the the contour plot ). By divergent term you mean : $\frac{i}{\pi^{2}r e^{i\phi}}$ ? Finally, I am having slightly different pre-factors in the final result. Your answer was very helpful for me ... Thanks a lot! Apr 12, 2021 at 14:02
• Thank you. This is good that you checked the result - I did only quick check ($I=0$ at $k=0$ and $a=0$). You are right: the factor $\frac{1}{2}$ just before the contour plot is extra (but, luckily, it is not used in the further calculations :) - I will remove it from the text. Divergent term - also correct. You are also right regarding the residual at $x=\frac{i}{2}$: $\,\sinh^2(\pi i/2)=-\sin^2(\pi/2)$ - I lost this sign. Sorry for these mistakes. Such calculations require attention and thoroughness... Apr 12, 2021 at 14:26
• Dear @Svyatoslav this is what I got as a final result : $\frac{i}{\pi} \frac{(2x-\pi)\sin(2a)+2a\sinh(2x)}{\cos(2a)+\cosh(2x)}$ and I hope I am not mistaken. Apr 12, 2021 at 15:57
• I'm afraid $I(x\to0)\to0$ - because of the symmetry (in fact we integrate $\sin kx$ from $e^{ikx}=\cos k x+i\sin kx$ in the nominator - integral with $\cos k x$ - wanish). Apr 12, 2021 at 16:49