Fourier series summation in David Griffiths' textbook

[sydfloyd]

Homework Statement

This isn't really a homework question, but something I've been wanting to know out of curiosity in David Griffiths' Introduction to Electrodynamics.

On pages 131 and 132, there is a Fourier series,

$$V(x,y) = \frac{4V_0}{\pi}\sum_{n=1,3,5...}\frac{1}{n}e^{\frac{-n \pi x}{a}}\sin{\frac{n \pi y}{a}}$$

The author then says that the series can be rewritten as

$$V(x,y) = \frac{2V_0}{\pi} \arctan{\frac{\sin{\frac{\pi y}{a}}}{\sinh{\frac{\pi x}{a}}}}$$

The author says "the infinite series ... can be summed explicitly (try your hand at it, if you like) ..." so I got curious and decided to take a shot at it.

I have been trying to figure out how to get this result all day, but after a few sheets of paper, I am still lost... Any help would be much appreciated.

P.S. I have not taken a course on complex analysis.

Homework Equations

$$\sin{u} = \frac{e^{iu}-e^{-iu}}{2i}$$
$$\sinh{u} = \frac{e^{u}-e^{-u}}{2}$$
$$\arctan{u} = \frac{i}{2}\ln{\frac{1-iu}{1+iu}}$$
$$\arctan{u} = \sum_{n=0}^{\inf} \frac{(-1)^n}{2n+1} u^{2n+1}$$

The Attempt at a Solution

 

[Ben Niehoff]

Most of these summations are done using clever applications of the Taylor expansion for the logarithm:

$$\ln (1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} z^n$$

Notice that your Fourier series has a 1/n in it. However, in your case, n is odd! So you'll have to play a bit with the formulas to get a series with only odd coefficients. Notice that

$$\ln (1+z^2)$$

will give a series with only even terms. Also, you can use Euler's formula

$$e^{a+ib} = e^a \cos b + i e^a \sin b$$

and so in particular, you can take the imaginary part:

$$\Im \left[ e^{a+ib} \right] = e^a \sin b$$

So why don't you play around with those things and see if you can sum the series.

 

[physdancer]

The solution is presented starting on page 3 of
http://academics.smcvt.edu/abrizard/EM/separation_two.pdf