Fourier transform of $\frac{\arctan(x)}{x}$

Question

If I try to compute the fourier transform of $$ f(x)=\frac{\arctan(x)}{x} $$ Maple return the function

$$ g(\omega) = \frac{1}{2} \pi e^{-|\omega|} $$

However when I try to compute the inverse fourier transform of $g$ I get

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

command I used are "fourier" and "invfourier". I was wondering if there's a way to derive by "hand" the fourier transform, the function is in $L^2$ so it exists.

Update:

Below a screenshot of the commands I used.

Answer 1

Write

$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$

so that the FT may be written as

$$\int_0^1 \frac{du}{u^2} \int_{-\infty}^{\infty} dx \frac{e^{i k x}}{\frac1{u^2}+x^2} $$

The inner integral is simply the FT of the Lorentzian function, or $\pi u e^{-|k|/u}$. The FT is then

$$\pi \int_0^1 \frac{du}{u} e^{-|k|/u} = \pi \int_1^{\infty} \frac{du}{u} e^{-|k| u} = \pi \Gamma(0,|k|)$$

Answer 2

Herein, we will use the form of the Fourier Transform of a function $f$ given by $$\mathscr{F}\left(f\right)=\int_{-\infty}^\infty f(x)e^{i\omega x}\,dx$$Integrals are interpreted in terms of Cauchy Principal Values.

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We can use the Convolution Theorem to find the Fourier Transform of $\frac{\arctan(x)}{x}$. First, we note that the Fourier Transforms of $\frac1x$ and $\arctan(x)$ are given respectively by

$$\mathscr{F}\left(\frac{1}{x}\right)=i\pi \,\text{sgn}(\omega) \tag 1$$

$$\begin{align} \mathscr{F}\left(\arctan(x)\right)&=\frac{i}{\omega}\,\mathscr{F}\left(\frac{d\arctan(x)}{dx}\right)\\\\ &=\frac{i}{\omega}\,\mathscr{F}\left(\frac{1}{1+x^2}\right)\\\\ &=\frac{i\,\pi}{\omega} \,e^{-|\omega|} \tag 2 \end{align}$$

Then, using the Convolution Theorem, we find that for $\omega\ne 0$, the Fourier Transform of $\frac{\arctan(x)}{x}$ is given by

$$\begin{align}\mathscr{F}\left(\frac{\arctan(x)}{x}\right)&=\text{PV}\left(\int_{-\infty}^\infty i\pi \,\text{sgn}(\omega-\omega')\,\frac{i\pi}{\omega'} \,e^{-|\omega'|}\,d\omega'\right)\\\\ &=-\frac{\pi}{2}\,\lim_{\epsilon\to0}\left(\int_{-\infty}^{-\epsilon}\text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'} \,d\omega'+\int_{\epsilon}^\infty \text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'}\,d\omega'\right)\\\\ &=-\frac{\pi}{2}\,\left(\int_{-\infty}^{-|\omega|}\text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'} \,d\omega'+\int_{|\omega|}^\infty \text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'}\,d\omega'\right)\\\\ &=\frac{\pi}{2}\,\left(-\int_{-\infty}^{-|\omega|}\frac{e^{\omega'}}{\omega'} \,d\omega'+\int_{|\omega|}^\infty \,\frac{e^{-\omega'}}{\omega'}\,d\omega'\right)\\\\ &=\pi \int_{|\omega|}^\infty \frac{e^{-\omega'}}{\omega'}\,d\omega'\\\\ &=\pi \Gamma\left(0,|\omega|\right) \end{align}$$

where $\Gamma(x,y)=\int_y^\infty u^{x-1}e^{-u}\,du$ is the upper Incomplete Gamma Function.