1
$\begingroup$

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!

$\endgroup$
5
1
$\begingroup$

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

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: enter image description here

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 $x\to 2x$) $$J=\frac{2i}{\pi}\,\frac{a\sinh 2x+x\sin 2a-\pi\sin a\sinh x}{\cosh 2x+\cos 2a}$$

Please check the calculations.

$\endgroup$
9
  • $\begingroup$ Just corrected several typos (lost $\frac{1}{\pi^2}$ in the first term - $\sinh^2\pi x=(\pi x)^2+...$ at $x\to0\,$) $\endgroup$
    – Svyatoslav
    Apr 11 at 17:10
  • $\begingroup$ 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! $\endgroup$
    – HJames
    Apr 12 at 14:02
  • $\begingroup$ 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... $\endgroup$
    – Svyatoslav
    Apr 12 at 14:26
  • $\begingroup$ 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. $\endgroup$
    – HJames
    Apr 12 at 15:57
  • $\begingroup$ 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). $\endgroup$
    – Svyatoslav
    Apr 12 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.