<s>[INST] How can I integrate this? (for calculate value of L-function ) I want to calculate the definite integral:
$$
\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx.
$$
Indeed, I already know that $\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx=L(\chi,1)=\frac{\pi}{\sqrt{5}}$ where $\chi$ is the Dirichlet character for $\mathbb{Z}[\sqrt{-5}]$.
I have some trouble with this actual calculation.
My calculation is as following:
\begin{eqnarray}
&&\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx \cr
&=& \int_{0}^{1} \frac{1+x^{6}}{1-x^{2}+x^{4}-x^{6}+x^{8}} dx\cr
&=& \int_{0}^{1} \frac{x^{2}(1+x^{2})(\frac{1}{x^{2}}-1+x^{2})}{x^{4}(\frac{1}{x^{4}}-\frac{1}{x^{2}}+1-x^{2}+x^{4})}dx
\end{eqnarray}
Substitute $x-\frac{1}{x}=t$. Then, $(1+\frac{1}{x^{2}})dx= dt$.
So The integral is
$$
\int_{-\infty}^{0} \frac{t^{2}+1}{t^{4}+3t^{2}+1}dt
$$
However, it has different value with the value which is calculated by the site 
http://www.emathhelp.net/calculators/calculus-2/definite-integral-calculator/?f=%28t%5E%7B2%7D%2B1%29%2F%28t%5E%7B4%7D%2B3t%5E%7B2%7D%2B1%29&var=&a=-inf&b=0&steps=on
So I've just given up to try more.
Is there any fine solution?
 [/INST] Noting that
$$ t^4+3t^2+1=(t^2+\frac{3}{2})^2-(\frac{\sqrt5}{2})^2,$$
and
$$ \frac{t^2+1}{t^4+3t^2+1}=\frac{5-\sqrt{5}}{10 \left(t^2-\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}-\frac{-5-\sqrt{5}}{10\left(t^2+\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}$$
we have
\begin{eqnarray}
&&\int_{-\infty}^0\frac{t^2+1}{t^4+3t^2+1}dt\\
&=&\frac{1}{10}\int_{-\infty}^0\left(\frac{5-\sqrt{5}}{t^2+\frac{3}{2}-\frac{\sqrt5}{2}}+\frac{5+\sqrt5}{t^2+\frac{3}{2}+\frac{\sqrt5}{2}})\right)dt\\
&=&\frac{1}{10}\left(\frac{5-\sqrt{5}}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}}\arctan(\frac{x}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}})+\frac{5+\sqrt{5}}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}}\arctan(\frac{x}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}})\right)\bigg|_{-\infty}^0\\
&=&\frac{1}{10}\left(\frac{5-\sqrt{5}}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}}+\frac{5+\sqrt{5}}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}}\right)\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt5}.
\end{eqnarray}
</s>