Showing $ \int_{\mathbb{R}^{2}_{+}} dt_1 dt_2 e^{-[t_1^2 + 2(-y-i0) t_1 t_2 + t_2^2]}= \frac{1}{2} \int_0^{\infty} \frac{dt}{t^2+2(-y-i0)t+1} $

Question

I am trying to derive some improper integral given as follows : \begin{align} \int_{\mathbb{R}^{2}_{+}} dt_1 dt_2 e^{-[t_1^2 + 2(-y-i0) t_1 t_2 + t_2^2]}= \frac{1}{2} \int_0^{\infty} \frac{dt}{t^2+2(-y-i0)t+1} \end{align} Here I guess $\mathbb{R}^{2}_{+}$ denotes the region $\{ (t_1, t_2)| 0\leq t_1, t_2 \leq \infty\}$. I have trouble with the first step

I know from twisted Mellin transformation

\begin{align} \frac{1}{A(p)^n} = \frac{1}{\Gamma(n)} \int_0^{\infty} du \phantom{1} u^{n-1} e^{-u A(p)} \end{align} Hence my first trial was using \begin{align} e^{-[t_1^2+2(-y-i0)t_1 t_2 + t_2^2]} = e^{-t_1^2[1+2(-y-i0)\frac{t_2}{t_1}+ \left(\frac{t_2}{t_1}\right)^2]} \end{align} But this seems to apply hard due to $[\cdots]$ having $t_1$ dependence.

How one can obtain the second formula?

Answer 1

Let $T=\frac{t_1}{t_2}$ and $t=t_2^2$, then $dt_1 dt_2 \left| \frac{\partial (T,t)}{\partial (t_1, t_2)} \right| = dT dt = 2 dt_1 dt_2$, so \begin{align} \int_{\mathbb{R}^{2}_{+}} dt_1 dt_2 e^{-[t_1^2 + 2(-y-i0) t_1 t_2 + t_2^2]} = \frac{1}{2} \int dT dt e^{-t[1+2(-y-i0) T + T^2]} = \frac{1}{2} \int_0^{\infty} \frac{dT}{T^2+2(-y-i0)T+1} \end{align} In the last step, we used Twisted Mellin transformation.

Answer 2

Just denote $t=t_1/t_2$ and use $dt_1dt_2=t_2dt_2 dt.$