Prove that $\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx = \sqrt {2\pi} e^{{-(2t)^2}/2} $.

Question

Why $\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx = \sqrt {2\pi} e^{{-(2t)^2}/2} $.

This value is needed in a problem of Hermite orthogonal functions. I have solved the original using this value but wandering how to prove it?

Answer 1

$$ \int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx-i\int_{-\infty}^{\infty} e^{{-x^2}/2} \sin{2tx}\:dx $$ Since $e^{{-x^2}/2} \sin{2tx}$ is odd $$ I(t)=\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx $$ Then \begin{align} I'(t)&=-2\int_{-\infty}^{\infty} xe^{{-x^2}/2} \sin{2tx}\:dx \\ &=2e^{{-x^2}/2} \sin{2tx}\Bigg|_{-\infty}^{\infty}-4t\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx \\ &=-4tI(t) \end{align} So $I(t)=Ce^{-2t^2}$. Since $$ I(0)=\int_{-\infty}^{\infty} e^{{-x^2}/2}\:dx=\sqrt{2}\int_{-\infty}^{\infty} e^{{-u^2}}\:du=\sqrt{2\pi} $$ We have $C=\sqrt{2\pi}$, and $$ \int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\sqrt{2\pi}e^{-2t^2} $$

Answer 2

Another method. First do it for $$ \int_{-\infty}^\infty e^{-x^2/2}dx = \sqrt{2\pi} $$ then complete the square to do it for $$ \int_{-\infty}^\infty e^{-x^2/2}e^{-2zx}dx = \sqrt{2\pi}\;e^{2z^2} $$ with real $z$. Then observe that both sides are analytic functions of the complex variable $z$ which agree for real $z$. Therefore they agree for all complex $z$.

added
"complete the square"
$$ -x^2/2-2zx = -x^2/2-2zx-2z^2+2z^2 =-(x+2z)^2/2+2z^2 $$ Thus, substitute $y=x+2z$ (with $z$ constant), $dy = dx$, to get $$ \int_{-\infty}^\infty e^{-x^2/2}e^{-2zx}dx =\int_{-\infty}^\infty e^{-(x+2z)^2/2}e^{2z^2}dx =e^{2z^2}\int_{-\infty}^\infty e^{-y^2/2}dy = e^{2z^2}\sqrt{2\pi} $$