Define $$f(s)= \int_1^{\infty}\psi(x)(x^\frac{s}{2}+ x^\frac{1-s}{2})\frac{dx}{x} $$ where $ \psi(x)= \sum_{n=1}^{\infty}e^{-n^2\pi x}$ is the Jacobi Theta function.
Claim- $f(s)$ is Entire
My Try :- $s=a+ib$ then $f(a+ib)=u(a,b)+iv(a,b)$ satisfy Cauchy Riemann Equations. How to proceed?
$$g(s)=\int_1^\infty \psi(x) x^s dx$$ converges for all $s$. $$\lim_{h\to 0} \frac{g(s+h)-g(s)}{h}- \int_1^\infty \psi(x) \log(x)x^s dx $$ $$=\lim_{h\to 0} \int_1^\infty \psi(x) \int_s^{s+h}\log(x) \frac{x^z-x^s}{h} dz dx$$ $$=\lim_{h\to 0} \int_1^\infty \psi(x) \int_s^{s+h}\log(x)\frac1{h} \int_s^z \log(x) x^u dudz dx$$
$$=\lim_{h\to 0} \int_1^\infty \psi(x) O(h \log^2(x) x^{1+s}) dx=0$$ Thus $$g'(s)=\lim_{h\to 0} \frac{g(s+h)-g(s)}{h}=\int_1^\infty \psi(x) \log(x)x^s dx $$ and $g$ is complex differentiable, for all $s$, it is holomorphic on the whole complex plane, by the Cauchy integral formula it is entire (analytic everywhere) and everywhere equal to its Taylor series at $s_0$.