We are given the function $p(r)$ as: $$p(r)=\int_{-\infty}^{\infty}\frac{1}{2i}k(S(k)-1)e^{ikr}dk$$
with $S(k)$ being an even function. What is the Fourier transform $P(k)$ of $p(r)$?
I have no idea how to deal with this complicated computation. Is there some help?
There is no special technique in that case; by applying the definition of Fourier transform, we get : $$ \begin{array}{rcl} P(k) &=& \displaystyle \int_\mathbb{R} p(r)e^{-ikr} \,\mathrm{d}r \\ &=& \displaystyle \frac{1}{2i} \int_\mathbb{R}\mathrm{d}r \int_\mathbb{R}\mathrm{d}k' \;k'(S(k')-1)e^{i(k'-k)r} \\ &=& \displaystyle -i\pi \int_\mathbb{R}\mathrm{d}k' \;k'(S(k')-1)\delta(k'-k) \\ &=& \displaystyle -i\pi k(S(k)-1) \end{array} $$ where we used the integral representation of the Dirac delta function, namely $\displaystyle \delta(k) = \int_\mathbb{R} \frac{\mathrm{d}r}{2\pi} \,e^{\pm ikr}$.