# Infinite series calculation of a potential

• I
Vitani1
TL;DR Summary
I'm having trouble beginning this...
I'm trying to get from the formula in the top to the formula in the bottom (See image: Series). My approach was to complexify the sine term and then use the fact that (see image: Series 1) for the infinite sum of 1/ne^-n. Then use the identity (see image: Series 2). Any other ideas?

#### Attachments

Let ##f(x)=\sum_{n=1}^\infty \frac{e^{-nx}}{n}##. Then ##f'(x)=\sum_{n=1}^\infty e^{-nx}=\frac{1}{e^x-1}##. Integrate to get ##f(x)=x-ln(e^x-1)##. When ##x=1##, we get ##f(1)=1-ln(e-1)##.

• Vitani1 and BvU
Fred Wright
Any other ideas?
Since you haven't included a homework statement and it's been awhile since your original post, I offer my solution to the problem.
We seek a closed form solution to the potential,
$$V(x,y)=\frac{4V_0}{\pi}S_n$$
$$S_n=\sum_{n=0}^{\infty}\frac{e^{-(2n+1)\frac{\pi}{a}x}\sin((2n+1)\frac{\pi}{a}y)}{(2n+1)}$$
I make the substitutions ##z=\frac{\pi}{a}(x+iy)## and ##\bar z=\frac{\pi}{a}(x-iy)## which scales the complex plane by ##\frac{\pi}{a}##. We now have
$$S_n=\frac{1}{2i}(\sum_{n=0}\frac{e^{-(2n+1)\bar z}}{(2n+1)}-\sum_{n=0}\frac{e^{-(2n+1) z}}{(2n+1)})$$
I observe
$$\frac{e^{-(2n+1)z}}{(2n+1)}=\int_{z}^{\infty} e^{-(2n+1) z'}d z'$$
and thus
$$\sum_{n=0}\frac{e^{-(2n+1)z}}{(2n+1)}=\sum_{n=0}\int_{z}^{\infty} dz'e^{-(2n+1)z'}$$
and interchanging the order of summation and integration we have
$$S_n=\int_{\bar z}^{\infty} d\bar{z'}\sum_{n=0}e^{-(2n+1)\bar{z'}}-\int_{z}^{\infty} dz'\sum_{n=0}e^{-(2n+1)z'}\\$$
The sum ##\sum_{n=0}e^{-(2n+1)z'}## can be evaluated as a geometric series
$$\sum_{n=0}e^{-(2n+1)z'}=e^{-z'}(1+e^{-2z'} + e^{-4z'} + ...)$$
$$=\frac{e^{-z'}}{1-e^{-2z'}}=\frac{1}{2\sinh(z')}$$
and we have
$$S_n=\frac{1}{2i}(\int_{z}^{\infty}\frac{d\bar{z'}}{2\sinh(\bar{z'})}-\int_{ z}^{\infty}\frac{d z'}{2\sinh(z')})$$
The upper limit of integration evaluates to ##\log(\tanh(\infty))=0## and we have
$$S_n=\frac{1}{4i}\left [\log(\tanh(\frac{ z}{2}))-\log(\tanh(\frac{\bar z}{2}))\right ]\\$$
The half hyperbolic tangent can be expressed as
$$\tanh(\frac{ z}{2})=\frac{(e^z-1)}{(e^z+1)}$$
and so
$$S_n=\frac{1}{4i}\left [\log(\frac{(e^z-1)(e^{\bar z}+1)}{(e^{\bar z}-1)(e^z+1)}) \right ]$$
$$=\frac{1}{4i}\left [ \log(\frac{\sinh(\frac{\pi}{a}x)+i\sin(\frac{\pi}{a}y)}{\sinh(\frac{\pi}{a}x)-i\sin(\frac{\pi}{a}y)}) \right ]$$
$$=\frac{1}{4i}\left [\log(-1) -\log(\frac{i-\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)}}{i+\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)}})\right ]$$
The ##\arctan(x)## can be expressed as
$$\arctan(x)=\frac{1}{2i}(\frac{i-x}{i+x})$$
and taking the principle branch of the logarithm we have
$$S_n=\frac{1}{2}\left [ \frac{\pi}{2}-\arctan(\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)})\right ]$$
$$V(x,y)=\frac{2V_0}{\pi}\left [\frac{\pi}{2}-\arctan(\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)})\right ]$$

• Vitani1 and Infrared