I am trying some questions from Apostol's Introduction to ANT and couldn't solve this particular problem on page 277.
I tried using Abel's summation Formula given in section 4.2 as follows:
$\frac{\chi(n) }{n^s}= \frac{S(M) }{M^s}+ s\int_{1}^{M}\frac{S(x) }{x^{s+1}} dx$ from which I got $L(s, \chi) = \sum_{n=M+1}^{\infty} \frac{\chi(n) }{n^s} + \frac{S(M) }{M^s}+s\int_{1}^{M}\frac{S(x) }{x^{s+1}} dx$
I am not able go to proceed from here as in integration limit is from 1 to $\infty$ and integrand is S(x) -S(N).
Can you please tell how should I proceed?
Thanks!!
For $\sigma>0$ and $1\le N<M$, by Abel's summation formula \begin{align} \sum_{n=1}^{M}\frac{\chi(n)}{n^s}&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}+\sum_{n=N+1}^{M}\frac{\chi(n)}{n^s}\\ &=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}-\frac{S(N)}{N^S}+\frac{S(M)}{M^S}+s\int_{N}^{M}\frac{S(x)}{x^{\sigma+1}}dx \end{align} Noting $\sigma>1$ and $S(x)=O(1)$, let $M\to\infty$. As a consequence $S(M)M^{-s}\to0$ and the integral converges. This means \begin{align} L(S,\chi)&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}-\frac{S(N)}{N^s}+s\int_{N}^{\infty}\frac{S(x)}{x^{\sigma+1}}dx\\ &=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}+s\int_{N}^{\infty}\frac{S(x)-S(N)}{x^{s+1}}dx \end{align} this completed the proof.