Calculating Fourier transform of the function $f(x)= \frac{x}{x^2+a^2}$

Question

How can one compute the Fourier transform in $L^2(\mathbb{R})$ of the function $f(x)= \frac{x}{x^2+a^2} \in L^2(\mathbb{R})$, where $a$ is some positive real constant? The Fourier transform in $L^1(\mathbb{R})$ is defined as $\hat{f}(t)= \int_{\mathbb{R}}f(x) e^{- i t x }dx$ and Fourier transform for $ f\in L^2(\mathbb{R})$ is defined as $$\hat{f}= l.i.m._{ n \rightarrow \infty} \int_{-n}^{n} f(x) e^{- i t x }dx$$

(It is limit in $L^2$-norm, as defined in book "Fourier and Wavelet Analysis" from authors Bachman & Narici).

Thanks a lot in advance.

Answer 1

We have $$f_a(x):={x\over x^2+a^2}={1\over 2i}\left [{1\over ix+a}+{1\over ix-a}\right ]$$ Observe that $$\int_0^\infty e^{-at}e^{itx}\,dt=-{1\over ix-a},\quad \int_{-\infty}^0 e^{at}e^{itx}\,dt={1\over ix+a}$$ Let $$g_a(t)=-{1\over 2i}{\rm sgn}\,(t)e^{-a|t|}={i\over 2}{\rm sgn}\,(t)e^{-a|t|}$$ By the Fourier inversion formula we get $$\widehat{f_a}(t)=2\pi g_a(t)=\pi i\,{\rm sgn}\,(t)e^{-a|t|}$$

Answer 2

I thought it might be instructive to present two distinct approaches to find the Fourier transform of $\frac{x}{x^2+a^2}$. To that end we now proceed.

METHODOLOGY 1: CONTOUR INTEGRATION

Let $f(x)=\frac{x}{x^2+a^a}$. Then, we have

$$\begin{align} \mathscr{F}\{f\}(k)&=\text{PV}\int_{-\infty}^\infty \frac{xe^{ikx}}{x^2+a^2}\,dx\\\\ &=2\pi i \begin{cases} \text{Res}\left(\frac{ze^{ikz}}{z^2+a^2}, z=i|a|\right)&, k>0\\\\ -\text{Res}\left(\frac{ze^{ikz}}{z^2+a^2}, z=-i|a|\right)&, k<0 \end{cases}\\\\ &=i\pi \text{sgn}(k)e^{-|ka|}\tag1 \end{align}$$

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METHODOLOGY 2: DIFFERENTIAL EQUATION APPLICATION

Let $g(x)=\frac{1}{x^2+a^2}$. Then, we have

$$\begin{align} G(k)&=\mathscr{F}\{g\}(k)\\\\ &=\int_{-\infty}^\infty \frac{e^{ikx}}{x^2+a^2}\,dx\\\\ &=\frac1{|a|}\int_{-\infty}^\infty \frac{e^{ik|a|x}}{x^2+1}\,dx\tag2 \end{align}$$

Owing to the uniform convergence of the integral of the differentiated integrand in $(2)$ for $k|a|$ bounded away from $0$ we have

$$\begin{align} G'(k)&=i\int_{-\infty}^\infty \frac{xe^{ik|a|x}}{x^2+1}\,dx\\\\ &=i\left(\text{PV}\int_{-\infty}^\infty \frac{e^{ik|a|x}}{x}\,dx-\text{PV}\int_{-\infty}^\infty \frac{e^{ik|a|x}}{x(x^2+1)}\,dx\right)\\\\ &=-\pi\text{sgn}(k)-i\text{PV}\int_{-\infty}^\infty \frac{e^{ik|a|x}}{x(x^2+1)}\,dx\tag3 \end{align}$$

Note that $\int_{-\infty}^\infty \frac{xe^{ikx}}{x^2+a^2}\,dx=-iG'(k)$.

Next, for $k|a|$ bounded away from $0$ we differentiate $G'(k)$ (uniform convergence applies again) in $(3)$ to find that

$$G''(k)=a^2G(k)\tag4$$

Solving the ODE of $(4)$ yields $G(k)=A^+e^{k|a|}+B^+e^{-k|a|}$ for $k>0$ and $G(k)=A^-e^{k|a|}+B^-e^{-k|a|}$ for $k<0$. Then, applying the initial conditions $G(0)=\frac\pi{|a|}$ and $G'(0^\pm)=\pm \pi$, we find that

$$G(k)=\frac{\pi}{|a|}e^{-{|ka|}}\tag5$$

Finally, differentiating $(5)$ and multiplying by $-i$ we find the coveted Fourier transform

$$\text{PV}\int_{-\infty}^\infty \frac{xe^{ikx}}{x^2+a^2}\,dx=i\pi \text{sgn}(k) e^{-|ka|}$$

which agrees with $(1)$.