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Recently I stumbled upon this short proof here:

$$L(1,\chi_2)=\sum_{j=0}^{+\infty}\left(\frac{1}{3j+1}-\frac{1}{3j+2}\right)=\int_{0}^{1}\frac{1-x}{1-x^3}\,dx=\int_{0}^{1}\frac{dx}{1+x+x^2}$$ so: $$\color{red}{L(1,\chi_2)}=\int_{0}^{1/2}\frac{dx}{x^2+3/4}=\frac{1}{\sqrt{3}}\arctan\sqrt{3}=\color{red}{\frac{\pi}{3\sqrt{3}}.}$$

So I was wondering: what if we have a polynomial $$Q(X)$$ in place of 1+x+x^2 in the denominator.Using partial fractions, the integral can be evaluated $$\sum_{j=0}^{\infty}a_n,$$ where $$a_n = \sum_{j=0}^{k}-\frac{1}{\alpha_i^n.n.Q'(\alpha_i)}$$

In the case of $$Q=x^2+x+1$$ we have that $$a_n/n$$ is actually a character mod 3, sending 1mod 3 to 1, 2 mod 3 to -1 and 0 mod 3 to 0.Can this be generalised?Also, we needn't stop at having just one polynomial, we can have a polynomial P at the numerator.If we can do this, we can evaluate L-functions as integrals of rational functions.

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Well, by the same approach shown in my previous answer, if $\chi$ is a non-principal character $\!\!\pmod{p}$ and $s\in\mathbb{N}^*$, we have that: $$ L(s,\chi)= \sum_{n\geq 1}\frac{\chi(n)}{n^s} = \int_{0}^{1}\frac{q_{\chi}(x)\log(x)^{s-1}}{1-x^{p}}\,dx $$ can be evaluated through partial fraction decomposition in terms of $\text{Li}_{s}$ evaluated at the $p$-th roots of unity and at one plus the $p$-th roots of unity. By the reflection formulas for the polylogarithms, in some cases $L(s,\chi)$ nicely simplifies. That happens for sure for $p=3$, since $1+\omega$ is a sixth root of unity, and if $\chi$ is the Legendre symbol, by Gauss sums.

For instance, if $\chi$ is the Legendre symbol $\!\!\pmod{5}$ and $s=2$, we have: $$\begin{eqnarray*}L(s,\chi)&=&\sum_{k\geq 0}\left(\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right)\\&=&-\sum_{k\geq 0}\int_{0}^{1}x^{5k}\log(x)(1-x-x^2+x^3)\,dx\\&=&-\int_{0}^{1}\frac{1-x-x^2+x^3}{1-x^5}\log(x)\,dx\\&=&-\int_{0}^{1}\frac{1-x^2}{1+x+x^2+x^3+x^4}\log(x)\,dx\end{eqnarray*} $$ and since $$ \int_{0}^{1}\frac{-\log(x)}{x-\zeta}\,dx = -\text{Li}_2\left(\zeta^{-1}\right) $$ by finding a partial fraction decomposition for $\frac{1-x^2}{1+x+x^2+x^3+x^4}$ through the residue theorem we get:

$$\begin{eqnarray*} L(s,\chi) &=& -\sum_{k=1}^{4} \text{Li}_2(e^{-2\pi i k/5})\,\text{Res}\left(\frac{1-x^2}{1+x+x^2+x^3+x^4},x=e^{2\pi i k /5}\right)\\&=&\frac{1}{25}\sum_{k=1}^{4}\left(\frac{k}{5}\right)\,\psi'\left(\frac{k}{5}\right)\end{eqnarray*}$$

that equals $\color{red}{\frac{4 \pi ^2}{25 \sqrt{5}}}$ by the properties of $\psi$ and $\psi'$.

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  • $\begingroup$ Im sorry if this is obvious, but what is $$q_{\chi}(x)$$?Im not sure how you found out that integral $\endgroup$ – Bogdan Simeonov Jun 10 '16 at 7:23
  • $\begingroup$ @BogdanSimeonov: I added a worked example. $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 11:45

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