Answer to Question #130825 in Complex Analysis for Kay

Question

Prove that for Re(s) >1,we have phi(s) =s integral from 1 to infinity f(x)/x^{s+1} dx

Answer 1

Let $f$ and $\phi(s)$ be given as in the theorem, and fix $s$ with finite abscissa of convergence $\sigma>min(0,\sigma_c)$ , so the Dirichlet series

$\phi(s)=\displaystyle\sum_{n=1}^\infty f(n)n^{-s}$

converges at $s$. Write

$f(x)=\displaystyle\sum_{n=1}^\infty f(n)\chi(x,n)$

where

$\chi(x,n)={1, if n\geqslant x\brace 0, if n<x}$

Then, for every $X\geqslant 1$

$s\intop_1^X \frac{f(x)}{x^{s+1}}dx=s\intop_1^X\displaystyle\sum_{n=1}^\infty\chi(x,n)\frac{f(n)}{x^{s+1}}dx=$ $s\intop_1^X\displaystyle\sum_{n\leqslant X}\chi(x,n)\frac{f(n)}{x^{s+1}}dx=$

$=s\displaystyle\sum_{n\leqslant 1}f(n)\intop_1^X\chi(x,n)\frac{1}{x^{s+1}}dx=$ $s\displaystyle\sum_{n\leqslant 1}f(n)\intop_1^n\frac{1}{x^{s+1}}dx=$ $=\displaystyle\sum_{n\leqslant 1}f(n)\frac{1}{s}(\frac{1}{n^s}-\frac{1}{X^s})=$ $\displaystyle\sum_{n\leqslant 1}\frac{f(n)}{n^s}-\frac{f(x)}{X^s}$

Now let $X->\infty$ . Then, by the convergence of $\phi(s)$, the first term on the right tends to $\phi(s)$. Moreover, Kronecker’s Lemma implies that the second term tends to 0. Hence we conclude

$lim_{X->\infty}s\intop_1^X \frac{f(x)}{x^{s+1}}=\phi(s)$.

which proves our formula. $\blacksquare$