Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 4

With the stated negative sign in the forward exponential, the inverse Fourier transform isFor example, this holds pointwise for Schwartz functions; more general square-integrable functions use the corresponding mean-square interpretation.

Differentiate under the Fourier transform integral:Hence the Fourier transform of a polynomial multiple isAbsolute integrability of the required moments justifies differentiation under the integral. The identity also extends to tempered distributions.

Let $A(k)$ be the Fourier transform of $e^{-x^2}$. Since $(e^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$, integration by parts gives $\mathcal{F}[x{e}^{-x^2}]=-(ik/2)A(k)$. Differentiation under the integral then yields $A^{\prime }(k)=-(k/2)A(k)$. The Gaussian integral supplies $A(0)=\sqrt{\pi }$, soApplying part (ii) with $n=1$ givesIt is imaginary and odd, as expected for the transform of a real odd function.

For sufficiently rapidly decaying functions, substitute the inverse Fourier transform of $g$ and interchange the integrals:The bracketed integral is the complex conjugate of $\tilde{f}(k)$. This is Parseval's identity for the chosen normalization. One may first prove it for Schwartz functions, where the interchanges are justified, and then extend it to $L^2$ by the Plancherel theorem.

Differentiating $\int e^{-a{x}^2}dx=\sqrt{\pi /a}$ with respect to $a>0$ gives $\int x^2{e}^{-a{x}^2}dx=\sqrt{\pi }/(2a^{3/2})$. ThereforeFrom part (iii), $|\tilde{f}(k){|}^2=(\pi /4)k^2{e}^{-k^2/2}$. Its frequency-side energy isin agreement with Parseval's identity.