You have a typo in your post. The first exponential in the answer given by Mathematica should read $e^{-2\pi w}$. In total the solution does NOT grow exponentially. The exponential growth of the $\Gamma$-function is "cured" by the factor $e^{-2\pi w}$.
To show that the integral does not grow exponential, it is possible to obtain an asymptotic expression for $w\to\infty$ without resorting to the explicit expression given in your post. To this end, we use integration by parts $\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx = \frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A + \frac{i}{2\pi w} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^2} dx.$ Repetitive integration by parts yields an asymptotic expansion for $w\to\infty$. The boundary term is the leading term, thus we have $\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right).$
For fun, I also give the next order term (if one is only interested in large $w$ this asymptotic expansion may prove more useful than the exact expression in terms of not so elementary functions). The next integration by parts yields $\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx=\frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A - \frac{e^{-2\pi iwx}}{[2\pi w(x-i)]^2} \Biggr|_{x=0}^A- \frac{1}{(2\pi w)^2} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^3} dx,$ which gives the next term in the asymptotic expansion $\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right) - \frac{1}{(2\pi w)^2} \left( 1+ \frac{e^{-2\pi i w A}}{ (A-i)^2}\right). $