Can one define a Fourier transform of $\log(a+|x|)$ with positive a?

Question

I would like to know if one can define the Fourier transform of $\log(a+|x|)$ with positive $a$ in some distributional sense. I tried to compute it using Mathematica, but the answer I get seems to be wrong since the inverse Fourier transform of it is not consistent. Does anyone have a trick to massage this function that allows the computation of the Fourier transform by hand? Thanks

Answer 1

Since tonight I'm on a roll: Per the discussion in the answer here we introduce a convergence factor in the integral and we consider

$$F_{\epsilon}(\omega)=\int_{-\infty}^{\infty}dx \log(|x|+a)e^{-i\omega x}e^{-\epsilon|x|}=e^{za}\int_{a}^{\infty}dx \log(x)e^{-zx}+ e^{z^* a}\int_{a}^{\infty}dx \log(x)e^{-z^*x}$$

where $z=\epsilon+i\omega$. Both integrals can be conveniently rewritten after an integration by parts as

$$e^{za}\int_{a}^{\infty}\log(x)e^{-zx}dx=\log a\frac{e^{za}}{z}+\frac{1}{z}\int_{a}^{\infty}e^{-zx}\frac{dx}{x}=\log a\frac{e^{-za}}{z}+\frac{e^{za}}{z}E_1(za)$$

where $E_1$ denotes the conventional exponential integral. It is well known that this function has only a log-type branch point at $z=0$ so if we are going to encounter interesting distributions in the $\epsilon\to 0$ limit, it is going to be at $\omega=0$. To isolate the distributional behavior at $\omega=0$ we exploit the series expansion of the special function around $x=0$ and we define $G(z)=\frac{E_1(z)+\gamma+\log(z)}{z}$, which is entire. The final result is

$$F_\epsilon(\omega)=\left((-\gamma-\log z)\frac{e^{za}}{z}+(-\gamma-\log z^*)\frac{e^{z^*a}}{z^*}\right)+a(e^{za}G(za)+e^{z^*a}G(z^*a))$$

which upon taking the limit gives (exercise)

$$F(\omega)=-2\pi\gamma\delta(\omega)-2\gamma\frac{\sin(\omega a)}{\omega}+2\pi\mathcal{P} \frac{\cos(a|\omega|)}{|\omega|}-2\mathcal{P}\frac{\sin(a|\omega|)\log|\omega|}{|\omega|}+a(e^{i\omega a}G(i\omega a)+e^{-i\omega a}G(-i\omega a))$$