# Value of (Fourier) series

1. Sep 14, 2015

### Incand

1. The problem statement, all variables and given/known data
Derive
$\sum_{n=1}^\infty \frac{1}{n^2+b^2} = \frac{\pi}{2b}\coth b\pi - \frac{1}{2b^2}$

from either
$e^{b\theta} = \frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{e^{in\theta}}{b-in}$
for $0 < \theta < 2\pi$.

or
$e^{b\theta} = \frac{\sinh b\pi}{\pi}\sum_{-\infty}^\infty \frac{(-1)^n}{b-in}e^{in\theta}$
for $-\pi < \theta < \pi$.

2. Relevant equations
N/A

3. The attempt at a solution
To get the series to be from $1\to \infty$ it seems a good idea to set $\theta=\pi$ which means we have to use the first equation. We get
$e^{b\pi } = \frac{e^{2\pi b}-1}{2\pi}\sum_{-\infty}^\infty \frac{e^{in\pi}}{b-in} =\frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{\cos (n\pi)(b+in)}{b^2+n^2}$
from that all $\sin n\pi = 0 \; \forall n \in Z$. Moving thing around and realising that $\cos (n\pi) = (-1)^n$ we get
$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} = \sum_{-\infty}^\infty \frac{(-1)^n(b+in)}{b^2+n^2} = b\sum_{-\infty}^\infty \frac{(-1)^n}{b^2+n^2}$
since the terms with $in$ all cancel our and are zero for $n=0$.

We realise that every term with $n$ is equal to the term with $-n$ and we get
$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_1^\infty \frac{1}{n^2+b^2}$
where we get $\frac{1}{b}$ from $n=0$. The sum of the series is then
$\sum_1^\infty \frac{1}{n^2+b^2} = \frac{\pi e^{b\pi }}{b(e^{2\pi b}-1)}-\frac{1}{2b^2}$
I can't get this to contain
$\coth b\pi = \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}}$.

2. Sep 15, 2015

### Dr. Courtney

How can you say the relevant equations are not applicable?

3. Sep 15, 2015

### Incand

They're not? The exercise give a warning "be careful" that may allude to that they or one of them may not be applicable. I thought since choosing $\theta = \pi$ seemed like a good idea to me and since $0 < \pi < 2\pi$ the first equation should be applicable?

Edit: I guess you meant why i wrote N/A. I just put them in the formulation of the exercise since they were that way in the book. I could've put them there instead ofcouse.

Last edited: Sep 15, 2015
4. Sep 15, 2015

### vela

Staff Emeritus
I don't see how both of these can be right because
$$\frac{e^{2\pi b}-1}{2\pi} = \frac{e^{\pi b}(e^{\pi b}-e^{-\pi b})}{2\pi} = \frac{e^{\pi b}}{\pi} \sinh \pi b,$$ so for $0 < \theta < \pi$, you'd have
$$e^{\pi b}\sum_{n=-\infty}^\infty \frac{e^{in\theta}}{b-in} = \sum_{n=-\infty}^\infty (-1)^n\frac{e^{in\theta}}{b-in}$$ Maybe that's true, but it doesn't look right to me. Consider the case when $b=0$, for instance.

You should have
$$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_{n=1}^\infty \frac{(-1)^n}{n^2+b^2}.$$ You changed the form of the summand when you dropped the alternating signs.

Last edited: Sep 15, 2015
5. Sep 16, 2015

### Incand

I'm pretty sure they're both right. I checked the book that I wrote them correctly (Folland, Fourier analysis and it's applications). It doesn't say that b is restricted in anyway but if we choose $b=0$ in the first equation we end up with $1 =0$ so perhaps $b \ne 0$. I think you end up with a special case when deriving the coefficients for $b=0$. Something along the lines of $c_n = \frac{1}{2\pi} \int_{-\pi}^\pi e^{inx}dx = \frac{-e^{-in\pi}+e^{in\pi}}{2in\pi} = \frac{1}{2\pi} \sin n\pi$. (complex sine function). So you get other $c_n$ when $b = 0$ but I only did this real quick so may be wrong. Actually thinking about this made me come up with the solution to the original problem, cheers!

Thanks! You(we) actually proved another execise in the book by mistake!

I think I got it now, posting the solution if anyone is interested. Since the functions are discontinuous I can't input $0$ or $\pi$ in the wrong equation to cancel our the alternatiing $(-1)^n$ and I have to use the convergence theorem:
If $f$ is $2\pi$-periodic and piecewise smooth we have (limit of partial sum)
$\lim_{N \to \infty} S_N^f (\theta ) = \frac{1}{2} \left[ f(\theta - ) + f(\theta +) \right]$
So If we pick $\theta = 0$ in the first equation we end up with
$\frac{e^{b2\pi}+e^{0}}{2} = \frac{e^{2\pi b}-1}{2\pi } \sum_{-\infty}^\infty \frac{b+in}{b^2+n^2}$
Rearanging we get
$b\sum_{-\infty}^\infty \frac{1}{n^2+b^2} = \frac{4\pi (e^{\pi b}+1)}{e^{2\pi b}-1}$
And summing from $1 \to \infty$ instead we get
$\frac{1}{b} + 2b\sum_1^\infty \frac{1}{n^2+b^2} = \frac{\pi (e^{2\pi b}+1)}{e^{2\pi b}-1}$
and we get
$\sum_1^\infty \frac{1}{n^2+b^2} = -\frac{1}{2b^2} + \frac{\pi (e^{2\pi b}+1)}{2b(e^{2\pi b}-1)} = -\frac{1}{2b^2} + \frac{\pi}{2b} \coth b\pi$

6. Sep 17, 2015

### vela

Staff Emeritus
$b=0$ was a poor choice because that would make $\sinh \pi b=0$, so I effectively divided by 0. I suspect with the first one, you end up having to take a limit as $b\to 0$ to evaluate the righthand side. While $\sinh \pi b \to 0$, the series probably diverges in such a way that the product is goes to 1.