Fourier transformation computation How can we compute the Fourier transform of the function:
$$
f(x) = \frac{1}{(1+x^2)^2}.
$$
Is the resulting function differentiable? If so, how many times?
I'm thinking if this would require partial fractions?
 A: I assume you know that the Fourier transform of $1/(1+x^2)$ is a constant multiple of $g(\xi)=e^{-|\xi|}$ (the exact form may depend on the definition of Fourier transform you are using). By the convolution theorem, $\hat f$ will be a constant multiple of
$$
g\ast g(\xi)=(1+|\xi|)e^{-|\xi|}.
$$
$\hat f$ is of class $C^2$ but not $C^3$. You do not need to compute $\hat f$ to know this. It is enough to realize that $|x|^2f\in L^1(\mathbb{R})$ but $|x|^3f\not\in L^1(\mathbb{R})$.
A: We are trying to compute
$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
&=\int_{-\infty}^\infty\frac{e^{2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\\
&=\int_{-\infty}^\infty\frac{t^3e^{2\pi ix}}{(t^2+x^2)^2}\mathrm{d}x\\
&=\int_\gamma\frac{t^3e^{2\pi iz}}{(t^2+z^2)^2}\mathrm{d}z\tag{1}
\end{align}
$$
where $\gamma$ is the contour that goes from $-L$ to $L$ along $\mathbb{R}$ then circles back counterclockwise to $-L$ through the upper half plane. Note that the size of the integrand on the circular part of the contour is $\le t^3/L^4$ and the length of the length of the circular part is $\pi L$, thus the integral over the contour goes to $0$ as $L\to\infty$. Also note that the integral does not depend on the sign of $t$, so we can assume $t>0$ and replace $t$ by $|t|$ in the final result.
By partial fractions, we get
$$
\begin{align}
\frac{t^3}{(t^2+z^2)^2}
&=\frac{t^3}{(z+it)^2(z-it)^2}\\
&=\frac{-t/4}{(z-it)^2}+\frac{-t/4}{(z+it)^2}+\frac{t/2}{(z-it)(z+it)}\\
&=\frac{-t/4}{(z-it)^2}+\frac{-t/4}{(z+it)^2}-\frac{i/4}{z-it}+\frac{i/4}{z+it}\tag{2}
\end{align}
$$
Puting $(1)$ and $(2)$ together yields
$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
&=\int_\gamma\frac{-t/4\;e^{2\pi iz}}{(z-it)^2}\mathrm{d}z
 +\int_\gamma\frac{-t/4\;e^{2\pi iz}}{(z+it)^2}\mathrm{d}z\\
&+\int_\gamma\frac{-i/4\;e^{2\pi iz}}{z-it}\mathrm{d}z
 +\int_\gamma\frac{i/4\;e^{2\pi iz}}{z+it}\mathrm{d}z\\[6pt]
&=2\pi i(-t/4)(2\pi i)e^{-2\pi t}+0\\[6pt]
&+2\pi i(-i/4)e^{-2\pi t}+0\\[6pt]
&=\pi^2t\;e^{-2\pi t}+\pi/2\;e^{-2\pi t}\tag{3}
\end{align}
$$
Thus, the final answer is
$$
\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
=\frac\pi2(1+2\pi|t|)\;e^{-2\pi|t|}\tag{4}
$$
Analysis of $\hat{f}$
$(\pi/2+\pi^2t)\;e^{-2\pi t}$ is $C^\infty(\mathbb{R})$. However, due to the discontinuity in the derivative of $|t|$ at $t=0$, care must be taken with the derivatives of $\hat{f}(t)=(\pi/2+\pi^2|t|)\;e^{-2\pi|t|}$ at $t=0$.
Taking the first and second derivatives of the formula in $(4)$ yields
$$
\frac{\mathrm{d}}{\mathrm{d}t}\hat{f}(t)=-2\pi^3t\,e^{-2\pi|t|}\to0\quad\text{as}\quad t\to0\tag{5}
$$
and
$$
\frac{\mathrm{d}^2}{\mathrm{d}t^2}\hat{f}(t)=-2\pi^3(1-2 \pi|t|)\,e^{-2 \pi|t|}\to-2\pi^3\quad\text{as}\quad t\to0\tag{6}
$$
However, the third derivative is discontinuous at $t=0$.
$$
\frac{\mathrm{d}^3}{\mathrm{d}t^3}\hat{f}(t)=8\pi^4(\rm{sgn}(t)-\pi t)\,e^{-2 \pi|t|}\tag{7}
$$
Another way to see that the first and second derivatives exist is that
$$
\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
=\int_{-\infty}^\infty\frac{-2\pi ix\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{8}
$$
and
$$
\frac{\mathrm{d}^2}{\mathrm{d}t^2}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
=\int_{-\infty}^\infty\frac{-4\pi^2x^2\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{9}
$$
converge absolutely.
However, the third derivative is not guaranteed to exist since
$$
\frac{\mathrm{d}^3}{\mathrm{d}t^3}\int_{-\infty}^\infty\frac{e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x
=\int_{-\infty}^\infty\frac{8\pi^3ix^3\,e^{-2\pi ixt}}{(1+x^2)^2}\mathrm{d}x\tag{10}
$$
does not converge absolutely.
A: By residue theory,
$$\forall t\in\mathbb{R}^+,\quad \int_{-\infty}^{+\infty}\frac{\cos(tx)}{(1+x^2)^2}\,dx = \frac{\pi}{2}\,(t+1)\,e^{-t},$$
so:
$$\int_{-\infty}^{+\infty}\frac{\cos(tx)}{(1+x^2)^2}\,dx = \frac{\pi}{2}\,(|t|+1)\,e^{-|t|}.$$
