Integral $\int_{0}^{A}\frac{\exp(-2\pi iwx)}{x-i}dx $ Here $A>0$, $w$-real,$\mathtt{i}$-complex.
Mathematica gives the answer:
$$\frac{1}{2}e^{2\pi w}(\mathtt{i}\pi+2\Gamma(0,2\pi w)-2\Gamma(0,2(1+\mathtt{i}A)\pi w)+2\ln(-\mathtt{i}+A)+2\ln(w)-2\ln(w+iA)) $$
My questions:
1. How to obtain this results without mathematica?
2. Why does this integral grow exponentially  as a function of $w$?
Mostly I don't understand why this integral explodes.The real part is:$$\int_{0}^{A}\frac{x\cos(2\pi wx)+\sin(2\pi wx)}{1+x^{2}}\ {d}x $$ and geometrically I don't understand why this integral grows so fast
 A: 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
$$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx  \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right) - \frac{1}{(2\pi w)^2} 
\left( 1+ \frac{e^{-2\pi i w A}}{ (A-i)^2}\right). $$
