Computing $ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}$ I would like to show that

$$ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}=\frac{\pi^2}{15}-\ln^2(\phi) $$

where $$ \phi = \frac{1+\sqrt{5}}{2}$$

How can this result be derived from the integral representation

$$  \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2} = -\int_0^{1/\phi^2}\frac{\ln(1-x)}{x} \mathrm dx $$

?
To justify the integral representation, observe that 
$$
\frac{-\ln(1-x)}{x}=\sum_{n\geq 1} \frac{x^{n-1}}{n}
$$
converges normally (hence uniformly absolutely) on $[0,1/\phi^2]$.
So term by term integration is possible and yields the desired formula.

Now to compute the integral, we need to know that an antiderivative of $-\ln(1-x)/x$ is the so-called dilogarithm. According to Wikipedia, Don Zagier once said that the dilogarithm is the only mathematical function with a sense of humor.

So 
$$
\sum_{n\geq 1} \frac{1}{\phi^{2n}n^2}=\mbox{Li}_2(1/\phi^2)=\mbox{Li}_2\left( \frac{3-\sqrt{5}}{2}\right).
$$

I found here: http://en.wikipedia.org/wiki/Polylogarithm that 
$$
\mbox{Li}_2\left( \frac{3-\sqrt{5}}{2}\right)=\frac{\pi^2}{15}-\ln^2\left(\frac{\sqrt{5}-1}{2}\right).
$$

That's your formula, since $\ln^2(1/\phi)=\ln^2(\phi)$.

The dilogarithm formula used above is most likely proved here: http://books.google.fr/books?id=u_UVn_iquj0C&amp;pg=PA141&amp;lpg=PA141&amp;dq=trilogarithm+identities&amp;source=bl&amp;ots=5lp7VWoN1p&amp;sig=xkvQm5jZFj_bcb0CpPKqsmIa1Fw&amp;hl=fr&amp;sa=X&amp;ei=d5EOUd_iPLK-0QH8k4CAAQ&amp;ved=0CDoQ6AEwAQ#v=onepage&amp;q=trilogarithm%20identities&amp;f=false
Use the fact that $- \ln (1-x) = \sum \frac{x^n}{n}$. Then prove you can swap sum and integral.