I am self studying number theory from Apostol's book and struck on this particular problem (Question 3 ).
$\Gamma(1-s) $is defined for every s and $\zeta(1-s) $ can also be defined for all s . So, F(a, s) is defined for all s.
Consider the equation F(a, s) = $\Gamma(1-s) $ ... in Question (2) .
Here$ F(a, s) =O(x^{-s}$ ). It is not analytic in $\sigma \leq 1$.
$\Gamma(1-s) $ is not analytic at s=1, 2,...,n,...( n belongs to integers ) and $\zeta(1-s) $ is not analytic at s=2.
So, F(a, s) is entire as for all s as s=1, 2 ,..., n are entire .
how to prove that it is entire?
I don't have any ideas except this .
Kindly guide.
For $0<a<1$, $$\sum_{n=1}^{N}e^{2\pi ina}=\frac{e^{2i\pi a}(e^{2\pi iaN})}{e^{2i\pi a}-1}=O(1)$$ Using this when applying Abel’s summation formula shows $$\sum_{n\le x}\frac{e^{2\pi ina}}{n^s}=O(x^{-s})$$ This means $F(s)$ is a convergent Dirichlet series for $\sigma > 0$, and hence is analytic there. Thus we only need to show the extended definition of $F(a, s)$ is analytic for $\sigma\le0$. Now in this region, $\Gamma(1 − s)$ is analytic everywhere and $\zeta(1 − s, ·)$ is analytic for $s\neq0$. This shows we only need to show $F(a, s)$ is analytic at $s = 0$ in order to show $F(a, s)$ is entire.
By Theorem 12.4, $\zeta(s, a)$ has a simple pole at $s = 1$ with residue $1$. Therefore there are entire functions $R_1(s, a)$ and $R_2(s, a)$ such that $$\zeta(1-s,a)=-\frac{1}{s}+R_1(s,a)\text{ and }\zeta(1-s,1-a)=-\frac{1}{s}+R_2(s,a)$$ Substituting shows there is an entire function $R_3(s, a)$ such that \begin{align} F(a,s)&=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\bigg{\{}-\frac{e^{\pi i(1-s)/2}+e^{\pi i(s-1)/2}}{s}+R_3(s,a)\bigg{\}}\\ &=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\bigg{\{}-\frac{2\sin(\pi s/2)}{s}+R_3(s,a)\bigg{\}} \end{align} We see $F(a, s)$ has a removable singularity at $s = 0$, hence by Riemann’s theorem on removable singularities $F(a, s)$ can be extended to an analytic function at $s = 0$.