Fourier transform of $H(x-1)/x$

Question

Consider $H(x-1)/x$ as a tempered distribution where $H$ is the Heaviside step function. I want to find an explicit form for its Fourier transform. Any ideas?

Answer 1

$$ \hat f(\nu)=\int_{-\infty}^{\infty}\frac{H(x-1)}{x}\mathrm e^{-i\nu x}\mathrm d x=\int_{1}^{\infty}\frac{\mathrm e^{-i\nu x}}{x}\mathrm d x=\Gamma(0,i\nu)=\mathrm E_1(i\nu)=-\mathrm{ci}(\nu)+i\,\mathrm{si}(\nu) $$ where $\Gamma(a,z)$ is the Incomplete Gamma function, $\mathrm E_1(x)$ is the $\mathrm E_n(x)$-Function for $n=1$ and $\mathrm{ci}(x)$ and $\mathrm{si}(x)=\mathrm{Si}(x)-\frac{\pi}{2}$ are the cosine integral and sine integral.