Fourier Transform of $\frac{1}{(1+x^2)^2}$

Question

I need to find the Fourier Transform of

$f(x) = \frac{1}{(1+x^2)^2}$

Where the Fourier Tranform is of $f$ is denoted as $\hat{f}$, where $\hat{f}$ is defined as $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$

I think I need to use the Fourier Inversion Theorem. From this theorem, I think we know that $\widehat{\widehat{\frac{1}{(1+x^2)^2}}} = (2\pi)\frac{1}{(1+x^2)^2}$

Then from the Fourier Inversion Theorem, I got that $\widehat{\frac{1}{(1+x^2)^2}} = \int_\mathbb{R}\frac{1}{(1+x^2)^2}e^{ixy}dx = \int_\mathbb{R}\frac{cos(xy)}{(1+x^2)^2}dx$

From here, I am unsure about how to calculate the integral, assuming I did everything right so far. Thanks for your help!

Answer 1

Method 1: Residues

Consider the contour integral

$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^2}$$

where $C$ is a semicircle in the upper half plane; here, $k>0$. Then by the residue theorem and Jordan's lemma:

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{i k\, e^{i k z}}{(z+i)^2} - \frac{2 e^{i k z}}{(z+i)^3} \right ]_{z=i}\\ &= \frac{\pi}{2} (k+1) e^{-k}\end{align}$$

For $k \lt 0$, $C$ is a semicircle in the lower half plane; for the same reasons as above, we have:

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z-i)^2} \right ]_{z=-i}\\ &= -i 2 \pi \left [\frac{i k\, e^{i k z}}{(z-i)^2} - \frac{2 e^{i k z}}{(z-i)^3} \right ]_{z=-i}\\ &= \frac{\pi}{2} (-k+1) e^{k}\end{align}$$

Therefore,

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi}{2} (1+|k|) e^{-|k|}$$

Method 2: Convolution

Knowing that the FT of $1/(1+x^2)$ is $\pi \, e^{-|k|}$, we may use the convolution theorem to deduce that

$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi^2}{2 \pi} \int_{-\infty}^{\infty} dk' e^{-|k'|} \, e^{-|k-k'|}$$

Again, the way we evaluate the integral on the RHS depends on the sign of $k$. For $k \gt 0$, this integral is

$$\frac{\pi}{2} \int_{-\infty}^0 dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{0}^k dk' \, e^{-k'} \, e^{-(k-k')}+ \frac{\pi}{2} \int_{k}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$

For $k \lt 0$, on the other hand, the integral is

$$\frac{\pi}{2} \int_{-\infty}^k dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{k}^0 dk' \, e^{k'} \, e^{k-k'}+ \frac{\pi}{2} \int_{0}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$

Evaluation of the above integrals reproduces the result derived above.

Answer 2

Tools: $$\mathcal F[f]=\hat f (\xi)=\int_{-\infty}^{+\infty}f(x)e^{-i\xi x} \mathrm dx $$ $$D \hat f=-i\widehat{xf} \tag{1}\label{1}$$ $$ \widehat{Df}=i\xi \hat f \tag{2}\label{2}$$ $$g(x)=\frac1{1+x^2}\quad\quad \hat g(\xi)=\pi e^{-|\xi|} \tag{3}\label{3}$$

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We want to compute the Fourier transform of $f(x)=\frac{1}{(1+x^2)^2}$. We begin with $g(x)=\frac1{1+x^2}$, noting that: $$\frac{\mathrm d}{\mathrm dx} \frac1{1+x^2}=-\frac{2x}{(1+x^2)^2} $$ namely $g'(x)=-2xf(x)$. By $\eqref{2}$, $\mathcal F[-2xf]=\mathcal F[g']=i\xi\hat g\stackrel{\eqref{3}}=i\xi\pi e^{-|\xi|}$. Hence: $$-2 \mathcal F[xf]=i\xi\pi e^{-|\xi|} $$ By $\eqref1$, $\mathcal F[xf]=i D \hat f$, so at the end we get: $$-2i D \hat f= i\xi\pi e^{-|\xi|}$$ Now we consider two different cases $\xi \ge 0$ and $\xi <0$ and we integrate: $$\hat f(\xi)=\frac{\pi}2(1+|\xi|)e^{-|\xi|}+c $$ Since $f\in L^1(\Bbb R)$, by the Riemann–Lebesgue lemma $\hat f(\xi)\to 0$ as $|\xi|\to \infty$, so we conclude that $c=0$. Finally: $$\hat f(\xi)=\frac{\pi}2(1+|\xi|)e^{-|\xi|}$$