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&pg=PA141&lpg=PA141&dq=trilogarithm+identities&source=bl&ots=5lp7VWoN1p&sig=xkvQm5jZFj_bcb0CpPKqsmIa1Fw&hl=fr&sa=X&ei=d5EOUd_iPLK-0QH8k4CAAQ&ved=0CDoQ6AEwAQ#v=onepage&q=trilogarithm%20identities&f=false
