I tried to proof this as a (signal engineering) homework using binomal square, but the example answer was given using differential equations. I'd like to know if my approach was possible.
I tried the following:
$$ t, v \in \mathbb{R}$$ $$\int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} dt $$ Expand with $e^{-\pi v^{2}} e^{\pi v^{2}}$
$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} e^{\pi v^{2}} dt $$ With $-v^{2} = (i v)^2$
$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi (t^{2} + 2 t (i v) + (iv)^{2})} dt $$
And then the binomal square
$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi (t + iv)^2} dt $$
Here I got stuck. I know $\int_{-\infty}^{\infty} e^{-\pi t^2} dt = 1$, but how to proof this is also true with a complex constant?
By parity, on has $$\int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} dt =\int_{-\infty}^{\infty} e^{-\pi t^{2}} \cos{(2 \pi v t)} \, dt.$$
Now, you can use the well-known formula $$\int_{-\infty}^\infty e^{-at^2}\cos(bt)\,dt =\frac{\sqrt\pi}{\sqrt{a}}e^{-\frac{b^2}{4a}}.$$ If you don't know this formula, see here.