ca.analysis and odes - An integral that somehow equals pi^2/6 and involves dilogarithms?

This problem can also be approached by rewriting the sum. I'll show a lot of details, probably too many, here.

Use the binomial series to write

$ frac{1}{(1+x^k)^2} = sum_{m=0}^{infty} binom{m+1}{1} (-1)^m x^{km} $

and insert this into the original sum $S$ to obtain

$ S = sum_{k=1} k^2 x^k sum_{m=0} (-1)^m (m+1) x^{km}. $

Interchange the sums to obtain

$ S = sum_{m=0} (m+1) (-1)^m sum_{k=1} k^2 x^{k(m+1)} $

$ = sum_{m=0} (m+1) (-1)^m frac{x^{m+1} (1+x^{m+1})}{(1-x^{m+1})^3} $

$ = -sum_{m=1} m (-1)^m frac{x^{m} (1+x^{m})}{(1-x^{m})^3}. $

Now use the factorization $1-x^m = (1-x)(1+x+x^2 +x^3 + cdots + x^{m-1}) $ to write

$ S = frac{-1}{(1-x)^3} sum_{m=1} (-1)^m m frac{x^m (1+x^m)}{(1+x+x^2 + cdots + x^{m-1})^3}. $

We have now factored out the most singular term as $x rightarrow 1.$ We can take the most rough approximation of the sum by finding its limit as $x rightarrow 1$, which gives

$ S approx frac{-1}{(1-x)^3} sum_{m=1} (-1)^m m frac{2}{m^3}.$

Using the known sum $ sum_{m=1} frac{ (-1)^m }{m^2} = -frac{pi^2}{12} $ now gives

$S approx frac{pi^2}{6} frac{1}{(1-x)^3} $ as desired....

NOTE: the last sum over $m$, if you don't just take $x=1$, but keep the powers $x^m$ in the numerator, gives dilogarithms of $x$ and $x^2$. However, I've probably thrown away too much in the denominator for that to be useful!!