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I'm having lots of troubles in computing this Fourier Transform:

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \ln(\beta x)\ e^{ikx}\ \text{d}x$$

Where $\beta\ \in\mathbb{R}$ and of course $\beta \neq 0$.

According to W. Mathematica, the result is the following:

$$-\frac{\sqrt{\frac{\pi }{2}}}{\left| k\right| }+\sqrt{2 \pi }\log (\beta) \delta (k)+\frac{i \pi ^{3/2} \delta (k)}{\sqrt{2}}-\gamma \sqrt{2 \pi } \delta (k)+\frac{\sqrt{\frac{\pi }{2}} }{k}$$

Can someone give me some hint?

I already tried to manipulate the integrand a bit, for example by(obviously) rewriting the log as a sum, and that made me to find the term

$$\sqrt{2 \pi } \log (\beta) \delta (k)$$

But for the remaining part I'm quite stuck, also because of the Euler-Mascheroni constant $\gamma$ that pops up somehow.

Thanks in advance!

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    What do you mean by the logarithm of a negative number? – 2017-02-02
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    $\alpha$ is really irrelevant in this question. Please keep the number of parameters as small as possible: the simpler, the better - this holds for answers and questions just as well. $\beta$ can also be removed through a suitable substitution. – 2017-02-02
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    @JackD'Aurizio You're right about $\alpha$, indeed I could have omitted it. – 2017-02-02
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    And you should clarify how $\log(x)$ is defined for $x<0$: is it just $\pi i+\log(-x)$? – 2017-02-02
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    @Edited for $\alpha$. And yes, for the logarithm we have $\pi i + \log(-x)$ – 2017-02-02
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    i would take laplace transforms and take the limit to vanishing real parameter in the end – 2017-02-02
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    Using the standardbranch of log, we see that the integral as it stands is not uniquely defined on the negative real axis. To fix this issue we understand the path of integration as $(-\infty +i\epsilon,+i\epsilon)$ in this region ($\epsilon\rightarrow0_+$). Furthermore, since $|\log(x)|>|x|^{-1}$ for $|x|\rightarrow\infty$ the integral furthermore doesn't exist in a usual sense but has to be interpreted in the sense of a proper limit (a.k.a in the sense of tempered distributions) This said, we might write formally... – 2017-02-02
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    ... $$ F_x\{\log(x)\}(k)=\lim_{\sigma\rightarrow0_+}\lim_{\epsilon\rightarrow0_+}\int_{-\infty+i\epsilon}^{i\epsilon}\log(x)e^{ikx+\sigma x}+\lim_{\sigma\rightarrow0_+}\int_0^{\infty}\log(x)e^{ikx-\sigma x} $$ or $$ F_x\{\log(x)\}(k)=\lim_{\sigma\rightarrow0_+}\left(\int_{-\infty}^0(\log(|x|)+i \pi)e^{ikx+\sigma x}+\int_0^{\infty}\log(x)e^{ikx-\sigma x}\right)=\\\lim_{\sigma\rightarrow0_+}\Re\int_0^{\infty}(\log(x)+i\pi)e^{i kx-\sigma x}=\lim_{\sigma\rightarrow0_+}\Re L_{x}\{\log(x)+i\pi\}(\sigma-ik) $$ – 2017-02-02
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    $\texttt{Mathematica 10.0.0.0}$ yields $$ -\frac{\pi + \,\mathrm{i}\left[\log\left(-\beta\right)-\log\left(\beta\right)\right] \text{sgn}\left(k\right)}{\,\sqrt{\,2 \pi\, }\, \left\vert\, k\,\right\vert } $$ – 2017-02-02
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    @FelixMarin Interesting and strange. My output was.. well the one above, instead. – 2017-02-02
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    @AlanTuring This is my code $\verb@Clear[\[Beta], k, x];@$ $$\verb@ Integrate[Log[\[Beta] x] Exp[I k x], {x, -Infinity, Infinity}, Assumptions -> Element[k, Reals] && (\[Beta] != 0)]/Sqrt[2 Pi]@ $$ – 2017-02-02
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    @FelixMarin Oh you did write it as an integral. I used the "FourierTransform" command! – 2017-02-02

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