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I have following integration $$I=\int_{0}^{\infty}\frac{\exp (i a x)}{(b x+1)^2}dx$$ where $i=\sqrt{-1}$ (for the complex notation).

I checked the book: Gradshteyn and Ryzhik -Table of integrals, Series and Products, in which I could not find any matching.

When I use Mathematica, it gives an answer with the incomplete Gamma function, which may be correct.

However, I really want to know how I can solve this integration step by step (at least come up with integration which is available in a book). So I tried integration by parts, and after two steps I have ended up with another integration which has form $$J=\int_{0}^{\infty}\exp (i a x)\ln(b x+1)dx$$ which does not have any readily available answer in that book.

Can someone please guide me to solve this integration?

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    If the bounds went from $-\infty$ to $\infty$ This would be a textbook example for residue theorem, and would have an easy answer. However, because it starts at 0, it is not quite so easy. I don't see a transformation that makes it simple. – 2017-02-21
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    It can be expressed in terms of a [Generalized Exponential Integral](http://dlmf.nist.gov/8.19) which is, of course, related to the Incomplete Gamma Function. – 2017-02-21

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We assume $b>0$ so that the integral converges (if $b<0$, it has a singularity at $x=1/b$). Setting $y=a(x+1/b)$, the bottom limit transforms to $1/b$, and the integral then becomes $$ I= \frac{e^{-a/b}}{b^2}\int_{a/b}^{\infty} \frac{e^{iy}}{y^2} \, dy $$ Lastly, we put $y=iz$, and deform the contour to return to the real axis, which reduces $I$ to the standard form $$ \frac{-ie^{-a/b}}{b^2}\int_{-ia/b}^{\infty} \frac{e^{-z}}{z^2} \, dz $$ An integration by parts puts this in the form of the exponential integral $\operatorname{Ei}(t) = -\int_{-t}^{\infty} e^{-s}/s \, ds $, or you can write it in terms of the upper incomplete $\Gamma$-function $\Gamma(-1,-ia/b)$.

Alternatively, you can split the $e^{iax}$ into $\cos{ax}$ and $\sin{ax}$, integrate by parts and write the answer in terms of the sine and cosine integrals, but there's no elementary form in general.

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    Thanks @Chappers. I derived Mathematica answer: i) one time integration by parts and ii) wrote `exp(iax)=exp(-(-iax))`- this may be a invalid trick. So probably your steps are more logical. – 2017-02-21