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The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

With the stated angular-frequency convention, the inverse Fourier transform is $f(x)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }\tilde{f}(k)e^{ikx}dk$. Direct integration gives ${\tilde{f}}_0(k)={\int }_0^{\infty }{e}^{-(\alpha +ik)x}dx=1/(\alpha +ik)$. For $n\ge 1$, integration by parts has zero boundary terms and gives ${\tilde{f}}_n=n{\tilde{f}}_{n-1}/(\alpha +ik)$. ThereforeThis also follows from the Fourier transform of a polynomial multiple, since multiplication by $x$ corresponds to $i{\partial }_k$.

To prove Parseval's identity, substitute the inverse transform of $g$ into $\int f^{\ast }g$. Under the assumed integrability, Fubini's theorem permits interchange, and the inner integral is $\int f(x{)}^{\ast }{e}^{ikx}dx=\tilde{f}(k{)}^{\ast }$. HenceApply this with $f=g=f_n$. The spatial Gamma integral is ${\int }_0^{\infty }{x}^{2n}{e}^{-2\alpha x}dx=(2n)!/(2\alpha {)}^{2n+1}$. The transformed squared modulus is $(n!{)}^2/({\alpha }^2+k^2{)}^{n+1}$, giving the scaled even rational Parseval integral