Fourier transform

Remarks

The integral converges absolutely because $\abs{f(x) e^{-2\pi i x \cdot \xi}} = \abs{f(x)}$. So $\abs{\hat f(\xi)} \le \int \abs{f}$, and $\hat f$ is continuous by dominated convergence. By the Riemann–Lebesgue lemma, $\hat f(\xi) \to 0$ as $\abs{\xi} \to \infty$. Other common conventions are $\int f(x) e^{-i x \cdot \xi}\, dx$ and $(2\pi)^{-n/2} \int f(x) e^{-i x \cdot \xi}\, dx$; they change the constants in the formulas below.

With the convention above, if $f$ and $\partial_j f$ are integrable, then $\widehat{\partial_j f}(\xi) = 2\pi i \xi_j \hat f(\xi)$. If $\hat f$ is also integrable, the inversion formula $f(x) = \int \hat f(\xi) e^{2\pi i x \cdot \xi}\, d\xi$ holds for almost every $x$. Plancherel's theorem extends $\mathcal F$ to a unitary map of $L^2(\R^n)$, and duality extends it to tempered distributions.

Example: $f(x) = e^{-\pi \abs{x}^2}$ satisfies $\hat f = f$. For $n = 1$, completing the square gives $\hat f(\xi) = e^{-\pi \xi^2} \int_\R e^{-\pi (x + i\xi)^2}\, dx$, and the last integral equals $\int_\R e^{-\pi x^2}\, dx = 1$ by a contour shift.

Non-example: the constant function $f = 1$ on $\R$ is not integrable, and the integral $\int_\R e^{-2\pi i x \xi}\, dx$ diverges for every $\xi$. Its Fourier transform exists only as a tempered distribution, namely the Dirac delta at $0$.