Help proving $L(1, \chi) = \frac{ \pi}{3 \sqrt{3}} $

Question

I'm taking an introductory class to Fourier Analysis, and I'm having trouble proving the relation above. My class is working with Stein's Fourier Analysis, though the book isn't yielding any helpful hints.

My attempt so far follows a method outlined by Masha Vlasenko at Trinity College: https://www.maths.tcd.ie/~vlasenko/MA2316/assignment.pdf

I start with the Dirichlet function at s = 1

$ L(1, \chi) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n} = \sum^{N - 1}_{k=1} \alpha_{k} \sum_{k=1}^{\infty} \frac{\chi_{k}(n)}{n} $

Where $\alpha_{k} = (\chi, \chi_{k})$ are Gauss sums. Vlasenko then shows that

$ \sum_{k=1}^{\infty} \frac{\chi_{k}(n)}{n} = -\ln(1 - e^{\frac{2\pi i k}{N}})$

Giving a what I assume is a general formula for $L(1, \chi)$

$\sum_{k=1}^{N - 1}-\alpha_{k} \ln(1 - e^{\frac{2\pi i k}{N}})$

For $N = 3$, this evaluates to

$L(1, \chi) = -\alpha_{1} \ln(1 - e^{\frac{2 \pi i}{3}}) - \alpha_{2} \ln(1 - e^{\frac{4 \pi i}{3}}) $

This is where I'm stuck. How do I evaluate Gauss sums, and these complex logarithms? Any hints or help is greatly appreciated.

Answer 1

There is an elementary way of doing this:

$$L(1,\chi) = 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+ \cdots = \sum_{n=0}^\infty \left( \frac{1}{3n+1}-\frac{1}{3n+2}\right) = \int_0^1 \frac{1-x}{1-x^3}dx $$ which can be shown by using geometric series, then interchange of integral and series is justified by Tonelli's theorem. Then $$\int \frac{1}{1+x+x^2}dx = \frac{2}{\sqrt{3}} \arctan\left(\frac{1 + 2 x}{\sqrt{3}}\right)$$ gives $$L(1,\chi) = \frac{\pi}{3\sqrt{3}}$$

Answer 2

You didn't define $\chi$. $N=3$ suggests $\chi$ is the primitive character $\bmod 3$, ie. $\chi(3n)= 0,\chi(3n+1) = 1,\chi(3n+2) = -1$.

Then you meant for $\chi$ non-principal, ie. not the trivial character $1_{\gcd(n,N)=1}$ (otherwise the series diverges) $$L(1, \chi) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n} = \sum_{n=1}^{\infty} \frac{1}{n}\frac{1}{N} \sum^{N - 1}_{k=1} \alpha_{k}e^{2i \pi n k/N}=\frac{1}{N} \sum^{N - 1}_{k=1} \alpha_{k} \sum_{n=1}^{\infty} \frac{e^{2i \pi n k/N}}{n}\\=-\frac{1}{N} \sum^{N - 1}_{k=1} \alpha_{k} \log(1-e^{2i \pi k/N})$$ where $\alpha_k= \sum_{n=1}^N \chi(n) e^{-2i \pi nk/N}$ (the Gauss sums, ie. the discrete Fourier transform of $ \chi(n)$) and we used the Taylor series for $- \log(1-z)$.

We obtain $\alpha_1 = e^{2i \pi /3}-e^{2i \pi 2/3}= i \sqrt{3},\alpha_2 = e^{2i \pi 2/3}-e^{-2i \pi 4/3} = -i \sqrt{3}$ so that $$L(1,\chi) = \frac{-1}{3}i \sqrt{3} \log(1-e^{2i \pi /3}) -i \sqrt{3} \log(1-e^{2i \pi 2/3}))=\frac{\pi}{3 \sqrt{3}}$$

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Note : it is always the case that $\alpha_k = \overline{\chi(k)} \alpha_1$ for $\gcd(k,N)=1$, and iff $\chi$ is primitive, then this stays true for every $k$.