This is a continue of my previous question Integral $\int_{0}^{A}\frac{\exp(-2\pi iwx)}{x-i}dx $.
Here $A>0$, $w$-real, $c$-complex number.
I have realized, that I need a precise answer for any $w$, not only an approximation for big $w$
By definition incomplete Gamma function is:$\Gamma(0,x)=\int_{x}^{\infty}t^{-1}e^{-t}dt $ Why does the simple change of variable doesn't work? $\int_{0}^{A}\frac{e^{-2\pi iw(x-c+c)}}{x-c}dx=e^{-2\pi iwc}\int_{-2\pi iwc}^{2\pi iw(A-c)}\frac{e^{-t}}{t}dt=e^{-2\pi iwc}[\Gamma(0,-2\pi iwc)-\Gamma(0,2\pi iw(A-c))] $
(change of the variable $2\pi iw(x-c)=t $).
Instead, the correct result, which mathematica gives, is : $\begin{multline} e^{-2\pi iwc}[-\Gamma(0,2i(A-c)\pi w)+\Gamma(0,-2\pi icw)\\ +\log(A-c)- log(-c)-\log(i(A-c)w)+\log(-icw)] \end{multline} $ The part with logarithm is just a sum of corresponding arguments. For some $c$ this sum is zero, but not always.