Limit of integral with Parseval

Question

I am trying to evaluate the limit $$ \lim_{n \to \infty} \int_{-\infty}^{\infty} \frac{\sin(nt)}{\pi t}f(t) \,dt, $$ where $$ f(t)=e^{-t^2+2t}. $$ Since I am working with Fourier transforms I thought that is what should be utilized. I found $$ \frac{\sin(nt)}{\pi t} \xrightarrow{\mathcal{F}} \theta(t+n) -\theta(t-n),\;\;\; e^{-t^2+2t} \xrightarrow{\mathcal{F}} \sqrt{\pi}e^{-t^2/4+it+1}, $$ where $\theta$ denotes the Heaviside function. Thus we should have: $$ \lim_{n \to \infty} \int_{-\infty}^{\infty} \frac{\sin(nt)}{\pi t}f(t) \,dt = \frac{\sqrt{\pi}}{2\pi} \lim_{n \to \infty} \int_{-\infty}^{\infty} \left(\theta(t+n) -\theta(t-n)\right)e^{-t^2/4+it+1} \,dt, $$ but the $\theta$ part is nothing but a function that equals $1$ between $-n$ and $n$, and $0$ elsewhere. So we should have $$ \frac{\sqrt{\pi}}{2\pi} \lim_{n \to \infty} \int_{-n}^{n}e^{-t^2/4+it+1} \,dt = \frac{\sqrt{\pi}}{2\pi} \int_{-\infty}^{\infty}e^{-\frac{1}{4}(t-2i)^2} \,dt. $$ This looks like some gaussian integral but I am not sure how to evaluate it. Any input is appreciated!

Answer 1

Indeed, it is a gaussian:

$ \frac{\sqrt{\pi}}{2\pi}\intop_{-\infty}^{\infty}e^{-\frac{1}{4}\left(t-2i\right)^{2}}dt=\frac{2\cdot\sqrt{2\pi}\sqrt{\pi}}{2\pi}\cdot\frac{1}{\sqrt{2\pi}\cdot2}\intop_{-\infty}^{\infty}e^{-\frac{\left(t-2i\right)^{2}}{2\cdot\left(\sqrt{2}\right)^{2}}}dt=\frac{1}{\sqrt{2}}\underset{\text{=1}}{\underbrace{\intop_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\cdot2}e^{-\frac{\left(t-2i\right)^{2}}{2\cdot\left(\sqrt{2}\right)^{2}}}dt}} $

And the integral is 1 since you integrate over a density of $ \mathcal{N}(2i,\sqrt{2})$.