For all $\xi \in \mathbb{C}$ we have $e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$

Question

This is Exercise 2.4 in Stein & Shakarchi's Complex Analysis. Prove that for all $\xi \in \mathbb{C}$ $$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$$ They prove it for the real case, so I assume that I'm supposed to use that. All I can think to do is write $\xi=a+ib$, which gives a term $e^{2\pi i xa}$ in the integral which becomes 1, since the power is a real multiple of $2\pi$ and $i$. That leaves $$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{-2\pi xb}\ \mathrm{d}x$$ but I don't have any idea where to go from there, or if this is even a fruitful direction. Any hints would be appreciated.

Answer 1

Write $\xi=a+bi$. The exponent inside the integrand looks like

$$\begin{array}{ll} -\pi x^2+2\pi ix\xi & =-\pi x^2+2\pi ix(a+bi) \\ & =-\pi x^2+2\pi iax-2\pi bx \\ & = -\pi (x^2+2bx)+2\pi iax \end{array}$$

The real part has both an $x$ and an $x^2$ term, but we may complete the square:

$$ \begin{array}{l} = -\pi(x^2+2bx+b^2)+\pi b^2+2\pi iax \\ =-\pi(x+b)^2+\pi b^2+2\pi iax. \end{array} $$

With the substitution $u=x+b$ ($\Leftrightarrow x=u-b$) this becomes

$$ \begin{array}{l} =-\pi u^2+\pi b^2+2\pi ia(u-b) \\ =-\pi u^2+2\pi iau+[\pi b^2-2\pi iab]. \end{array}$$

Therefore

$$ \begin{array}{ll} \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2+2\pi ix\xi)\,\mathrm{d}x & \displaystyle =\exp(\pi b^2-2\pi iab)\int_{-\infty}^{\infty} \exp(-\pi u^2+2\pi iau)\,\mathrm{d}u \\ & \displaystyle =\exp(\pi b^2-2\pi iab)\exp(-\pi a^2) \\ & =\exp(-\pi(a+bi)^2) \\ & =\exp(-\pi \xi^2). \end{array} $$

Answer 2

Note that the integral $$ \int_{-\infty}^{\infty} e^{-\pi x^2}e^{-2\pi i x\xi}\,\mathrm{d}x $$ is the Fourier transform of $e^{-\pi x^2}$. Now simplify $$ e^{-\pi x^2}e^{-2\pi ix\xi} = e^{-\pi(x^2+2\pi i x)} $$ and complete the square of $x^2 + 2\pi i x$ to deduce that $$ x^2 + 2\pi i x = x^2 + 2\pi i x -\xi^2 + \xi^2 = (x+i\xi) + \xi^2. $$ Substituting, this gives $$ e^{-\pi(x^2 + 2\pi i x)} = e^{-\pi(x+i\xi)^2-\pi\xi^2} = e^{-\pi(x+i\xi)^2} e^{- \pi\xi^2} $$ and hence $$ \int_{-\infty}^{\infty} e^{-\pi(x^2+2\pi ix\xi)}\,\mathrm{d}x= e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x. $$ Making the substitution $u = x+i\xi$ so that $\mathrm{d}x = \mathrm{d}u$ then yields $$ e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x = e^{-\pi\xi^2}\int_{-\infty}^{\infty}e^{-\pi u^2}\,\mathrm{d}u= e^{-\pi\xi^2}\frac{\sqrt{\pi}}{\sqrt{\pi}} = e^{-\pi\xi^2}, $$ as was to be shown. Note that this shows that $$ \mathscr{F}(e^{-\pi x^2}) = e^{-\pi \xi^2}, $$ where $\mathscr{F}$ is the Fourier transform on $L^2(\mathbb{R})$.

Answer 3

This is, in my opinion, the most conventional solution that uses Cauchy's theorem. Hope this helps others like myself :)

Let $\gamma$ be a closed parallelogram, starting from the origin, consisting of 4 segments $\gamma_1, \gamma_2, \gamma_3, \gamma_4$ in clockwise orientation sequentially.

plotting of the curve $\gamma$

As a complex function $f(z) = e^{\pi z^2}$ is holomorphic in this simply connected region, we can apply Cauchy's theorem to get \begin{align*} 0 = \int_\gamma f(z)\, dz = \underset{(1)}{\underbrace{\int_{\gamma_1} e^{\pi z^2}\, dz}} + \underset{(2)}{\underbrace{\int_{\gamma_2} e^{\pi z^2}\, dz}} + \underset{(3)}{\underbrace{\int_{\gamma_3} e^{\pi z^2}\, dz}} + \underset{(4)}{\underbrace{\int_{\gamma_4} e^{\pi z^2}\, dz}} \end{align*}

For $(1)$, \begin{align*} \lim_{R \rightarrow \infty} (1) = \lim_{R \rightarrow \infty} \int_{-R}^R e^{\pi(ix)^2} i\, dx = i \int_{-\infty}^\infty e^{-\pi x^2} i\, dx = i \end{align*}

Assume $\xi = a + ib$, $(2)$ is bounded absolutely by \begin{align*} |(2)| &= \left| \int_{\gamma_2} e^{\pi z^2}\, dz \right| \\ &\leq \sup_{0 \leq t \leq 1} \left| exp(\pi (\xi t + iR)^2)\, dz \right| \cdot 1 \\ &=\sup_{0 \leq t \leq 1} exp(\pi ((ta)^2-(tb+R)^2))\, dz \\ \end{align*} which goes to $0$ as $R \rightarrow \infty$, so that $\lim_{R \rightarrow \infty} (2) = 0$. Similarly, $\lim_{R \rightarrow \infty} (4) = 0$.

For $(3)$, it can be written as \begin{align*} (3) = \int_R^{-R} exp(\pi(\xi + iy)^2) i\, dy = -i\int_{-R}^R exp(\pi\xi^2 -\pi y^2 + 2\pi i\xi y)\, dy = - i e^{\pi\xi^2} \int_{-R}^R e^{-\pi y^2} e^{2 \pi i \xi y}\, dy \end{align*}

Assembling $(1), (2), (3)$ gives, \begin{align*} i = i e^{\pi\xi^2} \int_{-R}^R e^{-\pi y^2} e^{2 \pi i \xi y}\, dy \end{align*} Dividing both sides with ${i e^{\pi\xi^2}}$ gives the desired answer.