Gamma integral in Dirichlet L-series

Question

I am studying Dirichlet L-series in Algebraic Number Theory by Neukirch (Chap VII, section 2). In order to define the completed L-series of a character $\chi$ it started considering the gamma integral $$\Gamma(\chi,s)=\Gamma\biggl(\frac{s+p}{2}\biggr)=\int_0^\infty e^{-y}y^{(s+p)/2}\frac{dy}{y}.$$ Then substitute $y\mapsto\pi n^2y/m$ to obtain $$\biggl(\frac{m}{\pi}\biggr)^{\frac{s+p}{2}}\Gamma(\chi,s)\frac{1}{n^2}=\int_0^\infty n^pe^{-\pi n^2y/m}y^{(s+p)/2}\frac{dy}{y}$$ and multiplying by $\chi (n)$ and summing over all $n\in\mathbb{N}$, we get $$\quad \biggl(\frac{m}{\pi}\biggr)^{\frac{s+p}{2}}\Gamma(\chi,s)L(\chi,s)=\int_0^\infty\sum_{n=1}^\infty\chi(n)n^pe^{-\pi n^2y/m}y^{(s+p)/2}\frac{dy}{y}.$$ Here switching the order of summation and integration is justified because $$\sum_{n=1}^\infty\int_0^\infty |\chi(n)n^pe^{-\pi n^2y/m}y^{(s+p)/2}|\frac{dy}{y}\leq\biggl(\frac{m}{\pi}\biggr)^{(\text{Re}(s)+p)/2}\Gamma\biggl(\frac{Re(s)+p}{2}\biggr)\zeta(\text{Re}(s))<\infty.$$

Now, I know that the series in the integral converges uniformly but I am not getting how this last inequality comes out: if someone can explain it to me, it will be very appreciated.

Answer 1

Substitute $u=\pi n^2/m y$: \begin{align*} \int_0^\infty|\chi(n)n^pe^{-\pi n^2y/m}y^{(s+p)/2}|\frac{dy}y&=\int_0^\infty n^pe^{-\pi n^2y/m}y^{(\operatorname{Re} s+p)/2}\frac{dy}y\\ &=n^p(\pi n^2/m)^{-(\operatorname{Re} s+p)/2}\int_0^\infty e^{-u}u^{(\operatorname{Re} s+p)/2}\frac{du}u\\ &=n^{-\operatorname{Re} s}(m/\pi)^{(\operatorname{Re} s+p)/2}\Gamma\left(\frac{\operatorname{Re} s+p}{2}\right) \end{align*} Now sum:$$ \sum_{n=1}^\infty \int_0^\infty|\chi(n)n^pe^{-\pi n^2y/m}y^{(s+p)/2}|\frac{dy}y=(m/\pi)^{(\operatorname{Re} s+p)/2}\zeta(\operatorname{Re}s)\Gamma\left(\frac{\operatorname{Re} s+p}{2}\right), $$ so this inequality is actually an equality.