# How to turn this sum into an integral?

I have been trying to find the closed form of this sum to no avail. It was suggested to me to try and turn this sum into an integral and solve it like that. However, I am confused as to how to do this.

If we have a circle of radius $r$ with an $n$-gon inscribed within this circle (i.e. with the same circumradius), we can find the difference of the areas using:

$$A_n =\overbrace{\pi r^2}^\text{Area of circle}-\overbrace{\frac{1}{2} r^2 n \sin \left(\frac{2 \pi}{n}\right)}^\text{Area of n-gon} =r^2\left(\pi-\frac{1}{2} n \sin \left(\frac{2 \pi}{n}\right)\right)$$

I want to find the following sum (starting with $n=3$, i.e. the $n$-gon is a triangle):

$$\Lambda=\sum_{n=3}^\infty A_n = r^2\sum_{n=3}^\infty \left(\pi -\frac{1}{2} n \sin \left(\frac{2 \pi}{n}\right)\right) = r^2 \lim_{k \rightarrow \infty} \left(\pi (k-3)-\frac{1}{2} \sum_{n=3}^k n \sin \left(\frac{2 \pi}{n}\right)\right)$$

Expanding the $\sin$ using its Taylor's Series, we have

$$\Lambda= \sum_{n=3}^\infty \pi - \frac{1}{2}n\left(\frac{2\pi}{n} - \frac{(\frac{2\pi}{n})^3}{3!} + \cdots \right) = \frac{1}{2}\sum_{n=3}^\infty \sum_{m = 1}^\infty (-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!n^{2m}}$$

I would like some help with turning this sum into an integral to try and solve it. Any other suggestions for solving this sum are welcome as well.

My previous question on this sum can be found here: Convergence and closed form of this infinite series?

-
You really, really, really should have included a link to the original question, which was math.stackexchange.com/questions/136944/… –  Gerry Myerson May 2 '12 at 0:21
Have you seen this? –  Bill Dubuque May 2 '12 at 0:43
@GerryMyerson Edited, you are right. –  Argon May 2 '12 at 1:50
@Bill: That's a great post! However, in the present case, substituting $1/(2m+1)!$ by its integral representation, summing the geometric series over $m$ and evaluating the integral just leads back to the original sum over $n$. Did you have something else in mind? –  joriki May 2 '12 at 7:07
You could write the sum over $n$ in terms of $\zeta(2m)$ and use this integral representation; then the sum over $m$ can be performed and yields a sum of (modified) Bessel functions; but I doubt that the resulting integral can be evaluated in closed form. –  joriki May 2 '12 at 7:48

I'll extend the sum to start at $n=1$ to avoid complications:
\begin{align} \sum_{n=1}^\infty\sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!n^{2m}} &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\sum_{n=1}^\infty n^{-2m} \\ &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\zeta(2m)\tag1 \\ &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\frac1{\Gamma(2m)}\int_0^\infty\frac{t^{2m-1}}{\mathrm e^t-1}\mathrm dt \\ &= \int_0^\infty\sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\frac1{\Gamma(2m)}\frac{t^{2m-1}}{\mathrm e^t-1}\mathrm dt \\ &= \int_0^\infty\frac{-2\pi}{t(\mathrm e^t-1)}\sum_{m=1}^\infty\frac1{(2m-1)!(2m+1)!}\left(-(2\pi t)^2\right)^m\mathrm dt \\ &= 2\pi\int_0^\infty\frac{\operatorname{ber}_2(2\sqrt{2\pi t})}{t(\mathrm e^t-1)}\mathrm dt\;,\tag2 \end{align}
where the order of the summations can be changed because the double sum converges absolutely, the order of summation and integration can be changed by the dominated convergence theorem, and $\operatorname{ber}_2(t)$ is a Kelvin function. Both $(1)$ and $(2)$ can be evaluated numerically, and the results coincide; dividing by $2$ and subtracting $2\pi$ yields the approximate value $7.41721$ for your sum.
I was notified, no worries. As for notation: all of Abramowitz & Stegun, the DLMF, the Wolfram Functions site, and a few of the other special function textbooks I'm familiar with use $\mathrm{ber}_k(x)$/$\mathrm{bei}_k(x)$ to denote the Kelvin functions, so I don't know where the capitalization in the wiki article came from. –  Guess who it is. May 2 '12 at 13:41