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
From the identity $$ Li_2(x)+Li_2(1-x)=\frac{\pi^2}{6}-(\log x)(\log(1-x)) $$ it follows: $$ Li_2(\phi^{-1})+Li_2(\phi^{-2}) = \frac{\pi^2}{6} - 2\log^2\phi. $$ From the identity $$ Li_2(1-x)+Li_2(1-x^{-1})=-\frac{1}{2}\log^2 x$$ it follows: $$ Li_2(-\phi^{-1})+Li_2(\phi^{-2}) = -\frac{1}{2}\log^2\phi.$$ From the identity $$ Li_2(x)+Li_2(-x) = \frac{1}{2} Li_2(x^2)$$ and the previous ones it follows that: $$ \frac{5}{2}Li_2(\phi^{-2}) = \frac{\pi^2}{6}-\frac{5}{2}\log^2\phi, $$ so: $$ Li_2(\phi^{-2}) = \frac{\pi^2}{15}-\log^2(\phi). $$ To prove the used identities, it is sufficient to derive them with respect to $x$ and consider that: $$ \frac{d}{dx}Li_2(x) = \frac{-\log(1-x)}{x},\quad \frac{d}{dx}\log^2{x}=\frac{2\log x}{x},\quad \frac{d}{dx}(\log x \log(1-x)) = \frac{\log(1-x)}{x}-\frac{\log x}{1-x},\quad Li_2(0)=0,\quad Li_2(1)=\zeta(2)=\frac{\pi^2}{6}.$$
Use the fact that $- \ln (1-x) = \sum \frac{x^n}{n}$. Then prove you can swap sum and integral.
