# L-function identity involving Hurwitz zeta fuction

by Zermelo's_Choice   Last Updated February 26, 2017 18:20 PM

I am asked to prove the following L-Function identity $$L(s,\chi)=\frac{1}{q^s}\sum_{\ell=1}^{q-1}\chi(\ell)\zeta(s,\frac{\ell}{q}),$$ where $\zeta(s,a)=\sum_{n=1}^\infty\frac{1}{(n+a)^s}.$

I have the general outline of the proof, but there is one piece of the derivation that is handwavey. Below is my proof and I've highlighted the area which I'm having trouble justifying.

Proof. Let $\chi(n)$ be a Dirichlet character modulo $q$. Define $$\delta_{\ell}(n)= \begin{cases} \varphi(q),& \text{if n\equiv\ell (q)}\\ 0,& \text{else} \end{cases}$$ By lemma 2.2 in chapter 8 of Stein and Shakarchi's Fourier Analysis, $$\delta_\ell(n)=\sum_\chi\overline{\chi(\ell)}\chi(n).$$ Then \begin{align} \sum_\chi \overline{\chi(\ell)}L(s,\chi)&=\sum_\chi \overline{\chi(\ell)}\sum_{n=1}^\infty\frac{\chi(n)}{n^s}=\varphi(q)\sum_{n\equiv \ell(q)}\frac{1}{n^s}. \end{align} If we factor out a $q$ from $n=kq+\ell$, then $\frac{n}{q}=k+\frac{\ell}{q}.$ Hence, \begin{align} \sum_\chi \overline{\chi(\ell)}L(s,\chi)=\varphi(q)\frac{1}{q^s}\sum_{k=1}^\infty\frac{1}{(k+\frac{\ell}{q})^s}. \end{align} If we multiply both of (3) sides by $\chi(\ell)$ and sum $\ell$ over $\varphi(q)=q-1$, we have \begin{align} \sum_{\ell=1}^{q-1}\chi(\ell)\sum_\chi \overline{\chi(\ell)}L(s,\chi)&=\varphi(q)\frac{1}{q^s}\sum_{\ell=1}^{q-1}\chi(\ell)\sum_{k=1}^\infty\frac{1}{(k+\frac{\ell}{q})^s}\quad (1)\\ \color{red}{\iff\sum_{\ell=1}^{q-1}\sum_\chi \chi(\ell)\overline{\chi(\ell)}L(s,\chi)}&=\varphi(q)\frac{1}{q^s}\sum_{\ell=1}^{q-1}\chi(\ell)\zeta(s,k+\frac{\ell}{q})\quad\quad (2)\\ \color{red}{\iff \varphi(q)L(s,\chi)}&=\varphi(q)\frac{1}{q^s}\sum_{\ell=1}^{q-1}\chi(\ell)\zeta(s,k+\frac{\ell}{q})\quad(3)\\ \iff L(s,\chi)&=\frac{1}{q^s}\sum_{\ell=1}^{q-1}\chi(\ell)\zeta(s,\frac{\ell}{q})\quad(4). \end{align} Where we used the orthogonality of Dirichlet characters in line $(3)$.

For the highlighted pieces, is it valid to move $\overline{\chi(\ell)}$ inside the sum over all characters mod $q$? Moreover, is it even true that $$\sum_{\ell=1}^{q-1}\chi(\ell)\sum_\chi\overline{\chi(\ell)}=\sum_\chi\chi(\ell)\overline{\chi(\ell)}?$$ If both of the of the equalities above are wrong, what should I consider?

Tags :