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I was studying a technique in Gradshyten and Ryzhik's Part 12 "Some Log Integrals" on how to integrate something like

$\displaystyle \int_{0}^{1}\frac{\ln(x)}{x^{2}-x+1}dx$ and came across a relation I am wondering about.

The paper uses the roots of the above quadratic in the denominator of the integrand to arrive at the sum.

They state that $\displaystyle \sin\left (\frac{k\pi}{3}\right)$ has period 6:

$\displaystyle \frac{\sqrt{3}}{2}, \;\ \frac{\sqrt{3}}{2}, \;\ 0, \;\ \frac{-\sqrt{3}}{2}, \;\ \frac{-\sqrt{3}}{2}, \;\ 0$. Then, they repeat.

These are used to equate: $\displaystyle \sum_{k=1}^{\infty}\frac{\sin\left (\frac{\pi k}{3}\right )}{k^{2}}$ to

$\displaystyle \frac{\sqrt{3}}{2}\sum_{k=0}^{\infty}\left(\frac{1}{(6k+1)^{2}}+\frac{1}{(6k+2)^{2}}-\frac{1}{(6k+4)^{2}}-\frac{1}{(6k+5)^{2}}\right)$

The digamma is then used to evaluate the sums.

I have no problem with the paper except how they equate this sin sum to the sum on the right side. Given $\displaystyle\sum_{k=0}^{\infty}\frac{1}{(ak+b)^{2}}$, they use a=6 because that is the period of the sin, and they use b=1,2,4,5 because those sin values, in that order, are not 0.

But, they do not explain how the sin sum can be written like the right side in the first place. Perhaps it is something I should be seeing.

I even used this method to evaluate $\displaystyle -2\int_{0}^{\frac{\pi}{4}}\ln(2\sin(x))dx$.

I was able to derive the correct sums by equating

$\displaystyle \sum_{k=1}^{\infty}\frac{\sin\left (\frac{\pi k}{2}\right )}{k^{2}}=\frac{1}{16}\left(\sum_{k=0}^{\infty}\frac{1}{(4k+1)^{2}}-\sum_{k=0}^{\infty}\frac{1}{4k+3)^{2}}\right)$ and arrive at the solution.

But, I do not know how one can prove that the sin sum is related to the rational sum.

Can anyone explain how these seemingly unrelated sums are in fact the same?.

Cheers.

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Could we have a link to this article/paper? It sounds interesting... :) –  anorton Nov 25 '12 at 18:18
    
I tried but it would not let me post it. If you Google "Part 12 Gradshyten and Ryzhik", it will come up. I will try again. –  Cody Nov 25 '12 at 18:21
    
    
The proof consists of the following steps: 1. Evaluate the integral in terms of the polylogarithms. 2. Expand the polylogarithm and simplify it as a sine sum. 3. Make a further simplification to obtain a series involving rational terms. 4. Evaluate this sum in terms of the polygamma functions. My question is, which step you have trouble understanding. –  sos440 Nov 25 '12 at 18:31
    
Hi SOS: I do not know how they get that rational sum from a sine sum. I suppose your step 3. –  Cody Nov 25 '12 at 19:02

1 Answer 1

up vote 3 down vote accepted

$$ \begin{array}{rcl} && \sum_{k=1}^{\infty}\frac{\sin\left (\frac{\pi k}{3}\right )}{k^{2}} \\ &=& \small\sum_{k=1}^{\infty}\left( \frac{\sin\left (\frac{\pi 6k}{3}\right )}{(6k)^{2}}+ \frac{\sin\left (\frac{\pi (6k+1)}{3}\right )}{(6k+1)^{2}}+ \frac{\sin\left (\frac{\pi (6k+2)}{3}\right )}{(6k+2)^{2}}+ \frac{\sin\left (\frac{\pi (6k+3)}{3}\right )}{(6k+3)^{2}}+ \frac{\sin\left (\frac{\pi (6k+4)}{3}\right )}{(6k+4)^{2}}+ \frac{\sin\left (\frac{\pi (6k+5)}{3}\right )}{(6k+5)^{2}} \right) \\ &=& \frac{\sqrt{3}}{2} \sum_{k=1}^{\infty}\left( \frac{1}{(6k+1)^{2}}+ \frac{1}{(6k+2)^{2}}- \frac{1}{(6k+4)^{2}}- \frac{1}{(6k+5)^{2}} \right) \end{array}$$

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Thanks Cassandrao. I know they have a series for each term of the period, but I did not sub the 6k+1, 6k+2, etc into the sin. It is so obvious now. I feel stupid. Thanks much. –  Cody Nov 25 '12 at 20:56

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