Why is $\int_{-\infty}^\infty e^{-\pi x(t+i\frac{n}{x})^2}dt=\int_{-\infty}^\infty e^{-(\sqrt {\pi x} t)^2}dt$?

Question

Why is $\int_{-\infty}^\infty e^{-\pi x(t+i\frac{n}{x})^2}dt=\int_{-\infty}^\infty e^{-(\sqrt {\pi x} t)^2}dt$? I feel like substituting $s=t+i\frac{n}{x}$ would work but then how would the limits of integration change? Usually I would substitute $t=\phi(u)$ where $\phi(u)$ is a real function, hence the limits of integration can be calculated. But how about $s$ in this case? Any ideas?

Answer 1

The idea for this is to use the residue theorem. You take the function $f=e^{\sqrt{\pi x}t}$ and integrate it on the complex rectangle with the sides $[-R,R]$, $[R,R+i\frac{n}{x}]$, $[-R+i\frac{n}{x},R+i\frac{n}{x}]$ and $[-R,-R+i\frac{n}{x}]$. As $f$ is holomorphic the integral over this rectangle is zero. Next up you want to consider $R\to\infty$ and show that the integral over $[-R,-R+i\frac{n}{x}]$ and $[R,R+i\frac{n}{x}]$ converges to $0$. Then the result follows, as the sum of the integrals is $0$ but the direction of the second is inversed to the first, hence they are equal.