On evaluating $\rho(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)P(sz)\,dz.$

77 Views Asked by At

How do you evaluate the following integral?

$$\rho(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)P(sz)\,dz.$$

$P(\cdot)$ is the prime zeta function. I got the integral from taking $\rho(s)=\sum_{\rho\in\Bbb P}\exp\big(-\rho^s\big)$ and using the Mellin transform to extend the domain of $\rho(s)$.

The series $\rho(s)$ converges for $\Re(s)>0.$

By definition:

$$\rho(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\sum_{p\in\Bbb P}p^{-sz}\,dz.$$

So I ended up summing up the residues of the integrand at the poles and got:

$$ \rho(s)=\Gamma \bigg(1+\frac{1}{s}\bigg)+\sum_{n=0}^\infty \frac{(-1)^n}{n!}P(-ns) $$

Is this about right? I think I messed up somewhere.