- Prove that for all \(\xi \in \mathbb { C }\) we have \(e ^ { - \pi \xi ^ { 2 } } = \int _ { - \infty } ^ { \infty } e ^ { - \pi x ^ { 2 } } e ^ { 2 \pi i x \xi } d x .\)
Solution
Define, for \(\xi\in\mathbb C\), \[ F(\xi)=\int_{-\infty}^{\infty}e^{-\pi x^2}e^{2\pi i x\xi}\,dx. \] For \(\xi\) in any compact subset of \(\mathbb C\), the integrand and all its \(\xi\)-derivatives are dominated by a function of the form \[ C_N(1+|x|)^N e^{-\pi x^2+C|x|}, \] which is integrable. Hence \(F\) is entire and differentiation under the integral sign is valid.
Differentiating once, \[ F'(\xi)=2\pi i\int_{-\infty}^{\infty}x e^{-\pi x^2}e^{2\pi i x\xi}\,dx. \] Since \[ \frac{d}{dx}e^{-\pi x^2}=-2\pi x e^{-\pi x^2}, \] integration by parts gives \[ \begin{aligned} F'(\xi) &=-i\int_{-\infty}^{\infty}\frac{d}{dx}(e^{-\pi x^2})e^{2\pi i x\xi}\,dx\\ &=i\int_{-\infty}^{\infty}e^{-\pi x^2}(2\pi i\xi)e^{2\pi i x\xi}\,dx\\ &=-2\pi\xi F(\xi), \end{aligned} \] because the boundary term vanishes by Gaussian decay.
Thus \[ \frac{d}{d\xi}\left(e^{\pi\xi^2}F(\xi)\right)=0, \] so \(e^{\pi\xi^2}F(\xi)\) is constant on \(\mathbb C\). At \(\xi=0\), \[ F(0)=\int_{-\infty}^{\infty}e^{-\pi x^2}\,dx=1. \] Therefore \[ \boxed{F(\xi)=e^{-\pi\xi^2}} \] for every \(\xi\in\mathbb C\).