Inverse fourier transform of $1/k$ from properties

Question

I am studying physics and I came across an exercise where I needed to obtain the inverse Fourier transform of $k^{-1}$. By applying the property: $$kG(k) \to -ig'(x)$$ And the Dirac's delta definition, I got: $$-ig'(x) = \frac{1}{\sqrt{2\pi}}\int e^{ikx}dk = \sqrt{2\pi}\delta(x) \Rightarrow g(x) = i \sqrt{2\pi}\Theta(x) = i \sqrt{2\pi} \frac{\sigma(x) + 1}{2}$$ Where $\Theta(x)$ is the Heaviside step function and $\sigma(x)$ is the sign function. However, this result does not seem to agree with what I get from WolframAlpha: $$g(x)= -i \sqrt{\frac{\pi}{2}}\sigma(x)$$ Where did I go wrong? Could it be when taking the primitive of the delta?

Answer 1

Note that $g'(x)=i\sqrt{2\pi}\delta(x)$ implies that

$$g(x)=i\sqrt{2\pi} H(x)+C$$

for any integration constant $C$. Now, inasmuch as $g(x)=-g(-x)$, we find that $C$ is given by $C=-\frac{i}{2}\sqrt{2\pi}$.

Thus, we have

$$\begin{align} g(x)&=i\sqrt{2\pi}\left(H(x)-\frac12\right)\\\\ &=i\sqrt{\frac{\pi}{2}}\text{sgn}(x) \end{align}$$

The reason for the difference in sign is that Wolfram Alpha uses $e^{-ikx}$ as the kernel of the inverse Fourier Transform.