# What does $L(n,\chi_4)$ mean?

I was reading some articles related to Euler sums and the Riemann zeta function, when I came across this definition:

$$L(n,\chi_4) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^n}$$

What is this function called and how is it related to the zeta function?

• $\chi_4(n) = (-1)^{(n-1)/2} 1_{2 |n-1}$ is the unique non-trivial Dirichlet character modulo $4$. It is completely multiplicative so that $L(s,\chi_4) = \sum_{n=1}^\infty \chi(n) n^{-s}=\prod_p \frac{1}{1-\chi_4(p) p^{-s}}$. Also $L(s,\chi_4)$ is meromorphic (and even entire) on the whole complex plane, and since $\chi_4$ is its own discrete Fourier transform, it has a functional equation similar to $\zeta(s)$. In one word, a Riemann hypothesis is conjectured for $L(s,\chi_4)$. Those Dirichlet L-functions are also the core of the proof of Dirichlet theorem in arithmetic progressions – reuns Jul 21 '17 at 21:52
• @Pickle You may also be interested by this parallel exposition of the explicit formulas for $\zeta$ and your Dirichlet L-function with a link to a table of the imaginary parts of the first zeros. – Raymond Manzoni Jul 22 '17 at 7:34
• @Pickle: your specific series is named a Dirichlet beta function and allows (for $n$ integer) a parallel between Euler and Bernoulli numbers. – Raymond Manzoni Jul 23 '17 at 11:41
• @RaymondManzoni Thanks a lot! I was actually wondering if there was a specific name for the case where the character is $\chi_4$. That's exactly what I was looking for! – Klangen Jul 24 '17 at 8:46
• @RaymondManzoni The sum in the Dirichlet Beta function goes from 0 to infinity, however here it goes from 1 to infinty... – Klangen Jul 24 '17 at 8:47

That author is writing $\chi_4$ for a certain "character", defined by $$\chi_4(n) = \begin{cases} 1, &n \equiv 1\pmod 4,\\ -1,&n\equiv 3\pmod 4,\\ 0, &\text{otherwise} \end{cases}$$ and then $$L(s,\chi_4) = \sum_{n=1}^\infty \frac{\chi_4(n)}{n^s} = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}$$ is the corresponding "$L$-function" of that character.
The simple connection with the zeta function is that, taking a different character $\chi(n) = 1$ for all $n$, we get $$L(s,\chi) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \zeta(s)$$
• Thank you for the answer. However, it is not very clear to me how the denominator goes from $n^s$ to $(2k+1)^s$ when applying $\chi_4$ to the numerator... – Klangen Jul 24 '17 at 8:45
• $\chi_4(n)=0$ for even $n$, so leave those terms out of the sum. For odd $n$, write $n = 2k+1$ so $\chi_4(2k+1)=(-1)^k$. – GEdgar Jul 24 '17 at 8:56