Evaluate $L(1, \chi) = \sum_{n=1}^\infty \frac{\chi_5(n)}{n},$ for $\chi$ mod $5$

Question

My HW question is:

Evaluate the series $$L(1, \chi_5) = \sum_{n=1}^\infty \frac{\chi_5(n)}{n},$$ where $\chi_5$ is the unique nontrivial Dirichlet character mod $5$.

My work is: \begin{align*} L(1, \chi_5) = \sum_{n=1}^\infty \frac{\chi_5(n)}{n} &= \frac{\chi_5(1)}{1} + \frac{\chi_5(2)}{2} + \frac{\chi_5(3)}{3} + \frac{\chi_5(4)}{4} + \frac{\chi_5(6)}{6} + \frac{\chi_5(7)}{7} + \cdots \\ &= 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + \frac{1}{9} + \frac{1}{11} + \cdots \\ &= \left(1 + \frac{1}{4} + \frac{1}{6} + \frac{1}{9} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \cdots\right) \\ &= \sum_{n=0}^\infty\frac{1}{5n + 1} + \frac{1}{5n + 4} - \frac{1}{5n + 2} -\frac{1}{5n + 3}. \end{align*} Now, we define $$f(x) = \sum_{n=0}^\infty\frac{x^{5n + 1}}{5n + 1} + \frac{x^{5n + 4}}{5n + 4} - \frac{x^{5n + 2}}{5n + 2} - \frac{x^{5n + 3}}{5n + 3}$$ and differentiate termwise to find that \begin{align*} f'(x) &= \sum_{n=0}^\infty x^{5n} + x^{5n + 3} - x^{5n + 1} - x^{5n + 2}\\ &= \frac{1}{1 - x^5} + \frac{x^3}{1 - x^5} - \frac{x}{1 - x^5} - \frac{x^2}{1 - x^5} \\ &= \frac{1 + x^3 - x - x^2}{1 - x^5} \end{align*} so \begin{align*} f(1) &= \int_0^1 \frac{1 + x^3 - x - x^2}{1 - x^5} \\ &= .43041 \quad \text{(using Wolfram-Alpha)}. \end{align*}

However, a direct computation in wolfram alpha gives $4\pi^2/25$ which is about 1.57.

Where did I go wrong?

Answer 1

Your work is good, but your use of Wolfram alpha is not as what you want is $L(5,3,1)$ which indeed gives your result (note that the way Wolfram Alpha works $L(5,1,s)$ is the $L$ of the trivial character evaluated at $s$, so $L(5,1,2)=(1-1/25)\zeta(2)=\frac{24 \pi^2}{6 \times 25}=\frac{4 \pi^2}{25}$

Then the character are indexed by generators so $L(5,2,s)$ computes the complex $L$ corresponding to the generator character $\chi_{5,2}$ and then $L(5,3,s)$ computes the $L$ for the unique real Dirichlet character.

Note that modulo $5$ there are $3$ nontrivial characters, but indeed only one is real.

Answer 2

When $\chi$ is a real primitive character modulo $q$, it is well known that

$$ L(1,\chi)=-{1\over q^{\frac12}}\sum_{1\le r\le q}\chi(r)\log\sin{r\pi\over q}\tag1 $$

if $\chi(-1)=1$ and

$$ L(1,\chi)=-{\pi\over q^{\frac32}}\sum_{1\le r\le q}r\chi(r)\tag2 $$

if $\chi(-1)=-1$. In the case of OP, $\chi_5$ is the unique real primitive character modulo $q=5$, we use (1) to conclude that

\begin{aligned} L(1,\chi_5) &=-{1\over\sqrt5}\log\left(\sin\frac\pi5\sin{4\pi\over5}\over\sin{2\pi\over5}\sin{3\pi\over5}\right) \\ &={2\over\sqrt5}\log{\sin(2\pi/5)\over\sin(\pi/5)}={2\over\sqrt5}\log\left(2\cos\frac\pi5\right)\approx0.43041 \end{aligned}