Prove by justifying all steps that for all \(\xi \in {\mathbb C}\) we have \(\displaystyle e^{- \pi \xi^2} = \int_{- \infty}^\infty e^{- \pi x^2} e^{2 \pi i x \xi} dx \; .\)
Hint: You may use that fact in Example 1 on p. 42 of the textbook without proof, i.e., you may assume the above is true for real values of \(\xi\).
Solution
Goal: Prove that for all \(\xi\in\mathbb C\), \[ e^{-\pi\xi^2}=\int_{-\infty}^{\infty}e^{-\pi x^2}e^{2\pi i x\xi}\,dx. \]
1. Define the transform:
1.1. For \(\xi=u+iv\), the integrand satisfies\[|e^{-\pi x^2}e^{2\pi i x\xi}|=e^{-\pi x^2-2\pi vx}=e^{-\pi(x+v)^2+\pi v^2}.\]
1.2. Hence\[F(\xi):=\int_{-\infty}^{\infty}e^{-\pi x^2}e^{2\pi i x\xi}\,dx\]is well-defined for all \(\xi\in\mathbb C\) and is entire by dominated convergence on compact sets.
2. Derive a differential equation:
2.1. Differentiate under the integral to get\[F'(\xi)=\int_{-\infty}^{\infty}(2\pi i x)e^{-\pi x^2}e^{2\pi i x\xi}\,dx.\]
2.2. Integrate by parts:\[ \begin{aligned} F'(\xi) &= -i\int_{-\infty}^{\infty}\frac{d}{dx}\left(e^{-\pi x^2}\right)e^{2\pi i x\xi}\,dx\\ &= -i\Big[e^{-\pi x^2}e^{2\pi i x\xi}\Big]_{-\infty}^{\infty} +\,2\pi i\xi\int_{-\infty}^{\infty}e^{-\pi x^2}e^{2\pi i x\xi}\,dx \\ &= -2\pi\xi F(\xi). \end{aligned} \]
3. Solve the ODE:
3.1. \(F'+2\pi\xi F=0\) has general solution \(F(\xi)=Ce^{-\pi\xi^2}\).
3.2. Evaluate at \(\xi=0\):\[F(0)=\int_{-\infty}^{\infty}e^{-\pi x^2}\,dx=1,\]so \(C=1\).
4. Conclusion:By uniqueness of solutions to this linear ODE on entire functions,\[F(\xi)=e^{-\pi\xi^2}\]for all \(\xi\in\mathbb C\). Q.E.D.