I need to calculate an inverse Fourier transform $\int_{-\infty}^{\infty} \bigl( \frac{1}{f+K} \bigr) \, \operatorname{rect}(Bf) \, e^{j 2\pi f x} \, df$,
where $K$ is a constant, and $B$ is the width of the rectangular box: $\operatorname{rect}(Bf) = \begin{cases}1 &, -B/2 \leqslant f \leqslant B/2\\ 0 &, \text{otherwise.} \end{cases}$.
I ended up at a stage where I need to calculate the convolution $ (e^{-j2\pi Kx} j\pi\operatorname{sgn}(x)) * \operatorname{sinc}(Bx)$.
Any hints? Thanks in advance.
Using the notation:
$$\Pi(t) = \begin{cases} 1 & |t| < \frac{1}{2} \\ 0 & |t| > \frac{1}{2} \end{cases}$$ Hint:
$$\begin{align*} I &= \int_{-\infty}^\infty \dfrac{1}{f+K}\Pi\left(\dfrac{f}{B}\right)e^{j2\pi xf}df\\ \\ &= \int_{-\frac{B}{2}}^{\frac{B}{2}} \dfrac{1}{f+K}e^{j2\pi xf}df\\ \\ &= e^{-j2\pi x K}\int_{-\frac{B}{2}}^{\frac{B}{2}} \dfrac{1}{f+K}e^{j2\pi x(f+K)}df\\ \\ u &= f+K\\ \\ \dots \end{align*}$$
Finish the $u$ substitution and do the integration by parts.