I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$

Question

I start with a integral in complex plane $$\oint_c \frac{e^{izx} e^{zy} dz}{z\cosh(za)}$$ where $c$ is a countour starting in $z = -R$ along the real axis and jumping the pole at origin and continuing to $z = R$ and closing above the real axis with a semicircle circulating all the poles. I got $$\int_{0}^{\infty} \frac{e^{ky}\sin(kx)}{k\cosh(ky)}dk = \frac{\pi}{2} + 2\pi\sum_{n=1}^{\infty} (-1)^{n+1} \exp\left\{\frac{-xn\pi}{2a}\right\} \frac{\cos{\frac{yn\pi}{2a}}}{n}$$ and now i have no idea.

Answer 1

We can dispose with the sums altogether by using a rectangular contour. I would express the integral as

$$\phi(x,y) = \frac{V}{\pi} x \int_0^1 du \, \int_{-\infty}^{\infty} dk \frac{\cos{k x u} \cosh{k y}}{\cosh{k a}} $$

assuming $|y| \lt a$. So consider the contour integral

$$\oint_{\gamma} dz \frac{\cos{z x u} \cosh{z y}}{\cosh{z a}} $$

where $\gamma$ is the rectangle with vertices $-R$, $R$, $R + i \pi/a$, $-R + i \pi/a$. The contour integral is then equal to

$$\int_{-R}^R dk \frac{\cos{k x u} \cosh{k y}}{\cosh{k a}} + i \int_0^{\pi/a} d\nu \frac{\cos{(R+i \nu) x u} \, \cosh{(R+i \nu) y}}{\cosh{(R+i \nu) a}} \\ + \int_R^{-R} dk \frac{\cos{(k+i \pi/a) x u} \cosh{(k+i \pi/a) y}}{\cosh{(k+i \pi/a) a}}+ i \int_{\pi/a}^0 d\nu \frac{\cos{(-R+i \nu) x u} \, \cosh{(-R+i \nu) y}}{\cosh{(-R+i \nu) a}}$$

As $R \to \infty$ the second and fourth integrals vanish because $|y| \lt a$. Also,

$$\cosh{(k+i \pi/a) a} = -\cosh{k a}$$ $$\cosh{(k+i \pi/a) y} = \cosh{k y} \cos{(\pi y/a)} + i \sinh{k y} \sin{(\pi y/a)} $$ $$\cos{(k+i \pi/a) x u} = \cos{k x u} \cosh{(\pi x u/a)} - i \sin{k u} \sinh{(\pi x u/a)} $$

Note that the integral over the imaginary part of the integrand will be zero as it is an odd function over a symmetric interval. Thus, the contour integral is

$$[1+\cos{(\pi y/a)} \cosh{(\pi x u/a)}]\int_{-\infty}^{\infty} dk \frac{\cos{k x u} \cosh{k y}}{\cosh{k a}} \\+ \sin{(\pi y/a)} \sinh{(\pi x u/a)}\int_{-\infty}^{\infty} dk \frac{\sin{k x u} \sinh{k y}}{\cosh{k a}}$$

We have a second integral that we didn't bargain for. We deal with it by considering the integral

$$\oint_{\gamma} dz \frac{\sin{z x u} \sinh{z y}}{\cosh{z a}} $$

which, using the same analysis as above, we find to be equal to

$$[1+\cos{(\pi y/a)} \cosh{(\pi x u/a)}]\int_{-\infty}^{\infty} dk \frac{\sin{k x u} \sinh{k y}}{\cosh{k a}} \\- \sin{(\pi y/a)} \sinh{(\pi x u/a)}\int_{-\infty}^{\infty} dk \frac{\cos{k x u} \cosh{k y}}{\cosh{k a}}$$

We apply the residue theorem by considering the respective residues at the pole $z=i \pi/(2 a)$ of each integrand. If we denote the cosine integral as $C$ and the sine integral as $S$, then the residue theorem produces the equations

$$[1+\cos{(\pi y/a)} \cosh{(\pi x u/a)}] C + \sin{(\pi y/a)} \sinh{(\pi x u/a)} S = \frac{2 \pi}{a} \cosh{\left (\frac{\pi x u}{2 a} \right )} \cos{\left (\frac{\pi y}{2 a} \right )} $$

$$\sin{(\pi y/a)} \sinh{(\pi x u/a)} C - [1+\cos{(\pi y/a)} \cosh{(\pi x u/a)}] S = \frac{2 \pi}{a} \sinh{\left (\frac{\pi x u}{2 a} \right )} \sin{\left (\frac{\pi y}{2 a} \right )} $$

I will spare the reader the algebra involved and simply present the solution we seek:

$$C = \int_{-\infty}^{\infty} dk \frac{\cos{k x u} \cosh{k y}}{\cosh{k a}} = \frac{2 \pi}{a} \frac{\cosh{\left (\frac{\pi x u}{2 a} \right )} \cos{\left (\frac{\pi y}{2 a} \right )}}{\cosh{\left (\frac{\pi x u}{a} \right )} + \cos{\left (\frac{\pi y}{a} \right )}}$$

Of course, we're not quite done because $C$ is not the integral we sought. Rather, we need to integrate $C$ with respect to $u$ to get that integral. Thus,

$$\begin{align}\phi(x,y) &= \frac{2 V}{a} x \cos{\left (\frac{\pi y}{2 a} \right )} \int_0^1 du \frac{\cosh{\left (\frac{\pi x u}{2 a} \right )} }{\cosh{\left (\frac{\pi x u}{a} \right )} + \cos{\left (\frac{\pi y}{a} \right )}} \\ &= \frac{2 V}{\pi} \cos{\left (\frac{\pi y}{2 a} \right )} \int_0^1 \frac{d\sinh{\left (\frac{\pi x u}{2 a} \right )}}{ \sinh^2{\left (\frac{\pi x u}{2 a} \right )}+ \cos^2{\left (\frac{\pi y}{2 a} \right )} } \\ &= \frac{2 V}{\pi} \left [\arctan{\frac{\sinh{\left (\frac{\pi x u}{2 a} \right )}}{\cos{\left (\frac{\pi y}{2 a} \right )}}} \right ]_0^1\end{align}$$

Thus,

$$\phi(x,y) = \frac{2 V}{\pi} \arctan{\left [\frac{\sinh{\left (\frac{\pi x}{2 a} \right )}}{\cos{\left (\frac{\pi y}{2 a} \right )}}\right ]} $$

ADDENDUM

It should be noted that $\phi$ is clearly a solution of Laplace's equation for $x \ge 0$ and $|y| \le a$. The boundary conditions appear to be $\phi(0,y) = 0$ and $\phi(x,\pm a) = V$. Thus, this solution may be checked using a conformal mapping of the BC's to the right-half plane using the transformation $w=\sinh{[\pi z/(2 a)]}$.

Answer 2

Assuming your summation is correct we have $$-\sum_{n=1}^\infty (-1)^ne^{-\frac{\pi n x}{2a}}\frac{e^{\frac{\pi n y}{2a}i}+e^{-\frac{\pi n y}{2a}i}}{2n}=-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}+\frac{\pi n y}{2a}i}}{n}-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}-\frac{\pi n y}{2a}i}}{n}.$$ This can be rewritten $$-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{\left(e^{-\frac{\pi x}{2a}+\frac{\pi y}{2a}i}\right)^n}{n}-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{\left(e^{-\frac{\pi x}{2a}-\frac{\pi y}{2a}i}\right)^n}{n}.$$ Using the series $-\log(1+z)=\sum_{n=1}^\infty (-1)^n\frac{z^n}{n}$, $|z|<1$, we obtain $$\frac{1}{2}\log\left(1+e^{-\frac{\pi x}{2a}+\frac{\pi y}{2a}i}\right)+\frac{1}{2}\log\left(1+e^{-\frac{\pi x}{2a}-\frac{\pi y}{2a}i}\right),$$ for $\left|e^{-\frac{\pi x}{2a}\pm\frac{\pi y}{2a}i}\right|=e^{-\frac{\pi x}{2a}}<1\implies \frac{\pi x}{2a}>0$. Converting to trigonometric functions, $$\frac{1}{2}\log 2+\frac{1}{2}\log\left(\cosh \left(\frac{\pi x}{2 a}\right)-\sinh \left(\frac{\pi x}{2 a}\right)\right)+\frac{1}{2}\log\left( \cosh \left(\frac{\pi x}{2 a}\right)+\cos \left(\frac{\pi y}{2 a}\right)\right).$$ Hence, your sum is $$\frac{\pi}{2}+\pi\log 2+\pi\log\left(\cosh \left(\frac{\pi x}{2 a}\right)-\sinh \left(\frac{\pi x}{2 a}\right)\right)+\pi\log\left( \cosh \left(\frac{\pi x}{2 a}\right)+\cos \left(\frac{\pi y}{2 a}\right)\right).$$ Simplifying gives $$\frac{\pi}{2}+\pi\log 2-\frac{\pi^2 x}{2a}+\log\left(\cos\left(\frac{\pi x}{2a}\right)+\cosh\left(\frac{\pi x}{2a}\right)\right).$$

Answer 3

I recently posted this same problem unaware that it had already been posted and responded to.

I am sorry, but integrating w.r.t k seems backwards to me. k is more like a constant or sum index. Thus, I am using the usual x.

$$\int_{0}^{\infty}\frac{\sin(ax)\cosh(bx)}{x\cosh(\frac{\pi}{2}x)}$$

$$\;\ $$

$$\begin{align}=\int_{0}^{\infty}\frac{\sin(ax)}{x}\cdot \frac{e^{bx}+e^{-bx}}{e^{\pi x/2}+e^{-\pi x/2}}dx\end{align}$$

$$\;\ $$

$$\begin{align}=\int_{0}^{\infty}\frac{\sin(ax)}{x}\cdot \frac{e^{bx-\frac{\pi}{x}x}+e^{-bx-\frac{\pi}{2}x}}{1+e^{-\pi x}}\end{align}dx$$

$$\;\ $$

$$=\begin{align}\int_{0}^{\infty}\frac{\sin(ax)}{x}\sum_{n=0}^{\infty}(-1)^{n}e^{-\pi xn}(e^{bx-\frac{\pi}{2}n}+e^{-bx-\frac{\pi}{2}n})dx\end{align}$$

$$\;\ $$

$$\begin{align}=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}\left(\frac{e^{-x(\pi n+\frac{\pi}{2}-b)}\sin(ax)}{x}+\frac{e^{-x(\pi n+\frac{\pi}{2}+b)}\sin(ax)}{x}\right)dx\end{align}$$

$$\;\ $$

The integral(s) here are rather famous Frullani-like integrals. Using the known result: $$\;\ $$ $\displaystyle \int_{0}^{\infty}\frac{e^{-xt}\sin(ax)}{x}dx=\tan^{-1}\left(\frac{a}{t}\right)$,

we have:

$$\begin{align}=\sum_{n=0}^{\infty}(-1)^{n}\tan^{-1}\left(\frac{a}{\pi n+\frac{\pi}{2}-b}\right)+\sum_{n=0}^{\infty}(-1)^{n}\tan^{-1}\left(\frac{a}{\pi n+\frac{\pi}{2}+b}\right)\end{align}$$

$$\;\ $$

Note that Ramanujan explored sums of this form in his First Notebook:

$$\begin{align}\tanh^{-1}\left(\frac{\sinh(a)}{\cos(b)}\right)=\Im \log\left(1+\frac{\sinh(a)}{\cos(b)}i\right)\end{align}$$

$$\;\ $$

$$\begin{align} =\Im\log\left[\left(1+\frac{ia}{\frac{\pi}{2}\pm b}\right)\prod_{n=1}^{\infty}\left(1-\frac{ia}{2\pi n-(\frac{\pi}{2}\pm b)}\right)\left(1+\frac{ia}{2\pi n+\frac{\pi}{2}\pm b}\right)\times \left(1-\frac{ia}{(2n-1)\pi+\frac{\pi}{2}\pm b}\right)\left(1+\frac{ia}{(2n-1)\pi -(\frac{\pi}{2}\pm b)}\right)\right]\end{align}$$

$$\;\ $$

$$=\sum_{n=0}^{\infty}(-1)^{n}\tan^{-1}\left(\frac{a}{\pi n+c}\right)=\tan^{-1}\left(\frac{\sinh(a)}{\sin(c)}\right)$$

$$\;\ $$

Only in this case, $c=\frac{\pi}{2}\pm b\to \sin(\frac{\pi}{2}\pm b)=\cos(b)$, thus the result:

$$\;\ $$

$$\tan^{-1}\left(\frac{\sinh(a)}{\cos(b)}\right)=\cot^{-1}\left(\frac{\cos(b)}{\sinh(a)}\right)$$

follows.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\phi\pars{x,y} = {2V \over \pi} \int_{0}^{\infty}{\sin\pars{kx}\cosh\pars{ky} \over k\cosh\pars{ka}}}\,\dd k}$

---

With $\ds{\alpha \equiv x/\verts{a}}$ and $\ds{\beta \equiv y/\verts{a}}$: \begin{align} &{\pi \over 2V\alpha}\,\phi\pars{x,y} \equiv \bbox[5px,#ffd]{\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\cosh\pars{k\beta} \over \cosh\pars{k}}\dd k} \\[5mm] = &\ \int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{2\sinh\pars{k}\cosh\pars{k\beta} \over \sinh\pars{2k}}\dd k \\[5mm] & = {\cal I}\pars{\alpha,1 + \beta} + {\cal I}\pars{\alpha,1 - \beta}\label{1}\tag{1} \end{align} where \begin{align} {\cal I}\pars{\alpha,z} & \equiv \int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\sinh\pars{kz} \over \sinh\pars{2k}}\dd k \\[5mm] = &\ \int_{0}^{\infty}\pars{{1 \over 2} \int_{-1}^{1}\expo{\ic k\alpha q}\,\dd q} {\expo{-k\pars{2 - z}} - \expo{-k\pars{z + 2}} \over 1 - \expo{-4k}}\dd k \\[5mm] = &\ {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty} {\expo{-k\pars{2 - z - \ic\alpha q}}\,\, - \expo{-k\pars{z + 2 - \ic\alpha q}} \over 1 - \expo{-4k}}\dd k\,\dd q \\[5mm] = &\ {1 \over 8}\int_{-1}^{1}\int_{0}^{\infty} {\expo{-k\pars{1/2 - z/4 - \ic\alpha q\,/4}}\,\,\,\, - \expo{-k\pars{z/4 + 1/2 - \ic\alpha q\,/4}}\quad \over 1 - \expo{-k}}\,\dd k\,\dd q \\[5mm] = &\ {1 \over 8}\int_{-1}^{1} \bracks{% \Psi\pars{{z \over 4} + {1 \over 2} - {\alpha\ic \over 4}\,q} - \Psi\pars{{1 \over 2} - {z \over 4} - {\alpha\ic \over 4}\,q}} \dd q \end{align}

where I used the A & S Table ${\bf\color{black}{6.3.22}}$ identity.

Then, \begin{align} {\cal I}\pars{\alpha,z} & \equiv \int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\sinh\pars{kz} \over \sinh\pars{2k}}\dd k \\[5mm] = &\ \left.{1 \over 8}\pars{{4 \over \alpha}\,\ic} \ln\pars{\Gamma\pars{z/4 + 1/2 - \ic\alpha q/4} \over \Gamma\pars{1/2 - z/4 - \ic\alpha q/4}} \right\vert_{\,q\ =\ -1}^{\,q\ =\ 1} \\[5mm] = &\ {\ic \over 2\alpha}\, \ln\pars{{\Gamma\pars{1/2 + z/4 - \ic\alpha/4} \over \Gamma\pars{1/2 - z/4 - \ic\alpha/4}}\, {\Gamma\pars{1/2 - z/4 + \ic\alpha/4} \over \Gamma\pars{1/2 + z/4 + \ic\alpha/4}}} \\[5mm] = &\ {\ic \over 2\alpha}\, \ln\pars{\sin\pars{\pi/2 - \pi z/4 - \ic\pi\alpha/4} \over \sin\pars{\pi/2 + \pi z/4 - \ic\pi\alpha/4}} \\[5mm] = &\ {\ic \over 2\alpha}\, \ln\pars{\cos\pars{\pi z/4 + \ic\pi\alpha/4} \over \cos\pars{\pi z/4 - \ic\pi\alpha/4}} \\[5mm] = &\ {\ic \over 2\alpha}\, \ln\pars{\cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4} -\ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4} \over \cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4} + \ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4}} \\[5mm] = &\ -\,{1 \over \alpha}\,\Im \ln\pars{\cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4} -\ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4}} \\[5mm] = &\ {1 \over \alpha}\, \arctan\pars{\tan\pars{\pi z \over 4}\tanh\pars{\pi\alpha \over 4}} \end{align} Then ( see (\ref{1}) ), \begin{align} {\pi \over 2V\alpha}\,\phi\pars{x,y} & \equiv \bbox[5px,#ffd]{\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\cosh\pars{k\beta} \over \cosh\pars{k}}\dd k} \\[5mm] & = {1 \over \alpha}\left\{% \arctan\pars{\tan\pars{\pi\bracks{1 + \beta} \over 4} \tanh\pars{\pi\alpha \over 4}}\right. \\[2mm] & \left.\phantom{{1 \over \alpha}\left\{\,\,\,\,\right.} + \arctan\pars{\tan\pars{\pi\bracks{1 - \beta} \over 4} \tanh\pars{\pi\alpha \over 4}}\right\} \\[5mm] & = {1 \over \alpha} \arctan\pars{\sinh\pars{\pi\alpha \over 2} \sec\pars{\pi\beta \over 2}} \\[5mm] & = {1 \over \alpha} \arctan\pars{\sinh\pars{\pi x \over 2\verts{a}} \sec\pars{\pi y \over 2\verts{a}}} \end{align} Finaly, \begin{align} \phi\pars{x,y} & = \bbx{{2V \over \pi} \arctan\pars{\sinh\pars{\pi x \over 2\verts{a}} \sec\pars{\pi y \over 2a}}} \\ & \end{align}