Riemann Zeta type integral

Question

We know the following identity to be true $$\int_0^\infty {\rm d}y\,y^{a-1}\sum_{n=1}^{\infty}e^{-yn^2}=\zeta(2a)\Gamma(a).$$ However I need to compute a similar integral $$\int_0^\infty {\rm d}y\,y^{a-1}\sum_{n=1}^{\infty}e^{-y(n^2+\alpha^2)}$$ Is there any identity known for this integral? How do I compute this?

Answer 1

$$I=\int_0^\infty y^{a-1}\sum_{n=1}^{\infty}e^{-y(n^2+\alpha^2)}dy$$ Due to the Fubini-Tonelli theorem we can interchange the sum and the integral:

$$I=\sum_{n=1}^{\infty}\int_0^\infty y^{a-1}e^{-y(n^2+\alpha^2)}dy$$

Expand $e^{-y(n^2+\alpha^2)}$:

$$e^{-y(n^2+\alpha^2)} = \sum_{j=0}^{\infty} \frac{(-y)^j (n^2+\alpha^2)^j}{j!}$$

Applying the Ramanujan's master theorem

If we denote:

$$\varphi(j) = (n^2+\alpha^2)^j$$

We have

$$I=\sum_{n=1}^{\infty}\int_0^\infty y^{a-1}e^{-y(n^2+\alpha^2)}dy = \sum_{n=1}^{\infty}\Gamma(a)\frac{1}{(n^2+\alpha^2)^a}= \Gamma(a)\sum_{n=1}^{\infty}\frac{1}{(n^2+\alpha^2)^a}$$

and the right hand converges when $\operatorname{Re}(a)>\frac{1}{2}$

Answer 2

This integral is the sum over $n \geq 1$ of $\int_{0}^{\infty}{dy y^{a-1}e^{-y(n^2+\alpha^2)}}=\int_0^{\infty}{dy \frac{y^{a-1}e^{-y}}{(n^2+\alpha^2)^a}}$, so it is $\Gamma(a)\sum_{n=1}^{\infty}{\frac{1}{(n^2+\alpha^2)^a}}$. I’m not sure there’s a simplification working for all $a$, although there is one residue theorem-based method to evaluate the sum when $a$ is an integer (see eg the accepted answer of Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$).