= Solution
With the stated angular-frequency convention, the <inverse Fourier transform> is $f(x)=(2\pi)^{-1}\int_{-\infty}^{\infty}\widetilde f(k)e^{ikx}dk$. Direct integration gives $\widetilde f_0(k)=\int_0^\infty e^{-(\alpha+ik)x}dx=1/(\alpha+ik)$. For $n\geq1$, <integration by parts> has zero boundary terms and gives $\widetilde f_n=n\widetilde f_{n-1}/(\alpha+ik)$. Therefore
$$
\boxed{\widetilde f_n(k)=\frac{n!}{(\alpha+ik)^{n+1}}.}
$$
This 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 $\int f^*g$. Under the assumed integrability, <Fubini's theorem> permits interchange, and the inner integral is $\int f(x)^*e^{ikx}dx=\widetilde f(k)^*$. Hence
$$
\int_{\mathbb R}f(x)^*g(x)dx=\frac1{2\pi}\int_{\mathbb R}\widetilde f(k)^*\widetilde g(k)dk.
$$
Apply 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>
$$
\boxed{\int_{-\infty}^{\infty}\frac{dk}{(\alpha^2+k^2)^{n+1}}=\frac{2\pi(2n)!}{(n!)^2(2\alpha)^{2n+1}}.}
$$
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