0
$\begingroup$

I don't know how to prove it:

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

I tried to imitate the EXAMPLE 1 on the page 42,and put it into a toy contour like this: enter image description here and take $\xi=x_{0}+iy_{0}$ instead of $i\xi$, but it makes no sense.

  • 0
    it is easier to just complete the square in the exponent – 2017-02-06
  • 0
    or to google: "Fouriertransform of a Gaussian" – 2017-02-06
  • 0
    @tired another way? – 2017-02-06

1 Answers 1

1

Since $-\pi x^2 + 2\pi ix\xi = -\pi(x - i\xi)^2 - \pi\xi^2$, we write

$$\int_{-\infty}^\infty e^{-\pi x^2}e^{2\pi ix\xi}\, dx = \int_{-\infty}^\infty e^{-\pi(x - i\xi)^2}e^{-\pi\xi^2}\, dx = e^{-\pi\xi^2} \int_{-\infty}^\infty e^{-\pi(x - i\xi)^2}\, dx$$

Let $\xi = a + bi$. Using the $u$-substition $u = x + b$, we obtain

$$\int_{-\infty}^\infty e^{-\pi(x - i\xi)^2}\, dx = \int_{-\infty}^\infty e^{-\pi(u - ai)^2}\, du$$

The last integral can be computed using the rectangular contour $\Gamma(R)$ with vertices $-R, R, R - ai$, and $-R - ai$. Let $f(z) = e^{-\pi z^2}$. By Cauchy's theorem, $\int_{\Gamma(R)} f(z)\, dz = 0$. The integrals along the vertical edges of $\Gamma(R)$ are bounded by a constant times $e^{-\pi R^2}$, which tend to $0$ as $R\to \infty$. Hence,

$$\int_{-\infty}^\infty e^{-\pi(u - ai)^2}\, du = \int_{-\infty}^\infty e^{-\pi u^2}\, du = 1$$

and the value of the original integral is $e^{-\pi \xi^2}$.