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?
 A: You may notice that, through the substitution $x=\frac{1}{z}$, we have:
$$ I=\int_{0}^{1}\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx = \int_{1}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz $$
hence:
$$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz = \frac{1}{4}\int_{\mathbb{R}}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz $$
and the last integral can be easily computed through the residue theorem. 
Let $Z$ be the set of the primitive $20$th roots of unity with positive imaginary part, i.e.: 
$$Z=\left\{\exp\left(\frac{\pi i }{10}\right),\exp\left(\frac{3\pi i }{10}\right),\exp\left(\frac{7\pi i }{10}\right),\exp\left(\frac{9\pi i }{10}\right)\right\}=\{\zeta_1,\zeta_2,\zeta_3,\zeta_4\}.$$
We have:
$$\begin{eqnarray*} I &=& \frac{2\pi i}{4}\sum_{j=1}^{4}\text{Res}\left(\frac{1+z^6}{1-z^2+z^4-z^6+z^8},z=\zeta_i\right)\\&=&\frac{2\pi i}{4}\left(-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}-\frac{i}{2\sqrt{5}}\right)\\&=&\color{red}{\frac{\pi}{\sqrt{5}}}\end{eqnarray*}$$
as wanted. Ultimately, that $\sqrt{5}$ just depends on a well-known quadratic Gauss sum (have a look at this Wikipedia entry, too), but the identity $L(\chi,1)=\frac{\pi}{\sqrt{5}}$ is also a consequence of Kronecker's formula, since the ring of integers of $\mathbb{Z}[\sqrt{5}]$ has class number one.
A: 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}
A: $$ I = \int_{-\infty}^{0} \frac{t^{2}+1}{t^{4}+3t^{2}+1}dt = \dfrac14\int_{-\infty}^\infty \frac{(t^{2}+it+1) + (t^{2}-it +1)}{(t^{2}+it+1)(t^{2}-it+1)}dt, $$
$$ I = \dfrac14\int_{-\infty}^\infty \frac{1}{t^{2}-it+1}dt + \dfrac14\int_{-\infty}^\infty \frac{1}{t^{2}+it+1}dt. $$
Choosing the poles in upper half-plane:
$$ I = \dfrac{\pi i}2\mathop{{\rm Res}}_{t=\frac{\sqrt5+1}{2}i}\frac{1}{t^{2}-it+1} + \dfrac{\pi i}2\mathop{{\rm Res}}_{t=\frac{\sqrt5-1}{2}i}\frac{1}{t^{2}+it+1},$$
$$ I = \dfrac{\pi i}2\lim_{t=\frac{\sqrt5+1}{2}i}\frac{1}{2t-i} + \frac{\pi i}2\lim_{t=\frac{\sqrt5-1}{2}i}\frac{1}{2t+i},$$
$$\boxed{I=\dfrac\pi{\sqrt5}}$$
