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Fourier transform of a Gaussian

Statement

For $t > 0$ and $\hat{f} (ξ) = \int_{R} f (x) e (- x ξ) d x$ , $\hat{e^{- π t x^{2}}} (ξ) = t^{- 1 / 2} e^{- π ξ^{2} / t} .$

Facts & Assumptions

Given: $t > 0$ ; the Gaussian is a Schwartz function.

Proof

technique · direct

1.1given

Put $G (ξ) = \int_{R} e^{- π x^{2}} e (- x ξ) d x$ . The Gaussian integral The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ , followed by $u = \sqrt{π} x$ , gives $G (0) = 1$ . On every compact $ξ$ -interval, the derivative of the integrand is dominated by a constant multiple of $∣ x ∣ e^{- π x^{2}}$ , so Differentiation under an improper multiple integral under an integrable derivative bound applied to real and imaginary parts gives $G^{'} (ξ) = \int_{R} (- 2 π i x) e^{- π x^{2}} e (- x ξ) d x .$

2.1step 1.1algebra

Since $(e^{- π x^{2}})^{'} = - 2 π x e^{- π x^{2}}$ , integration by parts on $[- R, R]$ and then $R \to \infty$ (the boundary term tends to $0$ ) yields $G^{'} (ξ) = - 2 π ξ G (ξ)$ . Hence $(e^{π ξ^{2}} G (ξ))^{'} = 0$ , and step 1.1 gives $G (ξ) = e^{- π ξ^{2}}$ .

3.1step 2.1algebra∎

Substitute $u = \sqrt{t} x$ in the defining integral and apply step 2.1 to obtain the stated $t^{- 1 / 2}$ scaling.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nickolas Andersen, Analytic Number Theory, Chapter 16 Gaussian transform (standard reference, not scraped)