Proving that $(2 \pi i)^{-1} \int e(\pi v^2/y^2) x^v y^{-1} dv = e(-\pi (\log x)^2 y^2 /4)$

Question

I've seen a particular integral transform (an inverse Mellin Transform) used a few times, but I don't know how it's proved. In particular, I'm trying to prove

$$\frac{1}{2\pi i} \int_{(2)} e^{\pi v^2/y^2} \frac{X^v}{y} dv = e^{-\frac{y^2}{4\pi}(\log X)^2},$$

where $X$ and $y$ are positive constants, and the integral along $(2)$ means the line $2 \pm i\infty$.

In particular, this is used in the classical theory of modular forms to concentrate integrals along designated coefficients of the associated Dirichlet series. Some of the first nontrivial bounds on the Fourier coefficients of modular forms came from Rankin and Selberg, who used this identity.

I don't actually need a proof, but I always feel better about using identities that I understand.

Answer 1

Let's see :

Define $\alpha=\pi/y^2$ Close your contour of integration to the left, so that we integrate around a half-circle centerd at two. Because our Integrand is analytic in the complex plane we can shift our contour to the origin \begin{align} I(\alpha,X)=\frac{\sqrt{\alpha}}{ i 2\pi^{3/2} }\int_{-i \infty}^{i \infty}e^{\alpha \nu^2+\log(X)\nu} \end{align} Rotate the contour now by $\pi/2$ (Allowed by analyticity) . \begin{align} I(\alpha,X)=\frac{\sqrt{\alpha}}{ 2\pi^{3/2} }\int_{-\infty}^{\infty}e^{-\alpha \nu^2+i\log(X)\nu} \end{align} This is indeed just the Fouriertransform of a Gaussian with frequency $\log(X)$

Can you take it from here?

Edit: i also think that your exponent in the end should look like $e^{-\frac{y^2}{4\pi}\log(X)^2}$ because the width should be inverse to that of the original Gaussian.