# Does a closed form exist for this summation?

How do I calculate $$\sum_{k=1}^\infty \frac{k\sin(kx)}{1+k^2}$$ for $0<x<2\pi$? Wolfram alpha can't calculate it, but the sum surely converges.

-
First thing to see is that S(x) is the derivative of an other sum, $T(x)= \sum_{k=1}^{\infty} \frac{-\cos(kx)}{1+k^2}$ – Matt B. Jul 14 '14 at 8:44
Wolfram Functions says that $\displaystyle\sum_{k=1}^{\infty}\dfrac{k\sin{(kx)}}{a^2+k^2}=\dfrac{\pi\sinh {(a(\pi-x))}}{2\sinh {(a\pi)}}$. But how to prove it? – Pkwssis Jul 14 '14 at 9:00
I think you can translate this sum in to a integral and then try to use the contour integral methods to derive this formula. – Nil Mal Jul 14 '14 at 9:11

Using known Fourier series expansion $$\sinh(at)=\frac{2\sinh \pi a}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}n\sin nt}{n^2+a^2},\quad t\in(-\pi,\pi)$$ (which can be easily proven simply by calculating the Fourier coefficients for this function) and noting that $(-1)^{n+1}\sin nt=\sin((\pi-t)n)$ we get the series in question taking $a=1$ and $t=\pi-x$: $$\sum_{n=1}^\infty\frac{n\sin nx}{n^2+1}=\frac{\pi\sinh(\pi-x)}{2\sinh \pi},\quad x\in(0,2\pi).$$

-
Very neat solution! – mike Jul 14 '14 at 9:22

We can write the sum as a linear combination of the LerchPhi function

$$\Phi(z,s,a) = \sum_{k=0}^{\infty} \frac{z^k}{(k+a)^s}.$$

First, we write the sum as (using partial fraction)

$$S = \sum_{k=0}^{\infty}\frac{e^{ikx}}{4i(k-i)} -\sum_{k=0}^{\infty} \frac{e^{-ikx}}{4i(k-i)}+ \sum_{k=0}^{\infty}\frac{e^{ikx}}{4i(k+i)}-\sum_{k=0}^{\infty}\frac{e^{-ikx}}{4i(k+i)}$$

$$\implies S =\frac{1}{4i}( \Phi(e^{ix},1,-i)- \Phi(e^{-ix},1,-i)+ \Phi(e^{ix},1,i)- \Phi(e^{-ix},1,i) ).$$

Note: this is a general technique which can handle more general series than the one in considdration.

-
I don't see how the original sum can make a real argument into a complex value. For example with $x=1$ your formula is equal to $-0.285478-0.2751 i$ – Kristoffer Ryhl Jul 14 '14 at 9:04
@Mhenni Benghorbal. I guess that a $\Sigma$ is missing for the fourth term – Claude Leibovici Jul 14 '14 at 9:07
@Darksonn: Make sure of your calculation. With $x=1$ the answer is $.5709561705$. – Mhenni Benghorbal Jul 14 '14 at 9:07
@Mhenni Benghorbal. Your solution is very elegant (may I say, as usual ?) ! Thanks for this good stuff ! Cheers :) – Claude Leibovici Jul 14 '14 at 9:10
Nice solution! i din't know much about the Lerch zeta function though.. – Pkwssis Jul 14 '14 at 9:21

You may use Mathematica. I tried Mathematica 7.0 and it gives:

Where $_2F_1$ is the hypergeometric function:

-
What is $_2F_1$ – Kristoffer Ryhl Jul 14 '14 at 8:54
@Darksonn: It is the hypergeometric function. – Mhenni Benghorbal Jul 14 '14 at 8:58
The expression is a real valued function the plot of which being quite nice and simple. – Claude Leibovici Jul 14 '14 at 9:05
Moreover, it appears valid for all $x\ne2\pi n,\;n\in\mathbb{Z}$, unlike that given in the accepted answer. But the accepted solution is way nicer. – Ruslan Jul 14 '14 at 13:33