How we can solve: $\int_0^\infty \frac{\exp(ib x) }{x+b}d x$ The above integral has a pole at $x=-b$. $b$ is real and $b\ge 0$. To my knowledge such integral are solved using residue theorem. But, I am not able to find suitable contour to solve this integral. 
 A: You can evaluate it in terms of the exponential integral
$$ {\it Ei} \left( 1,-i{b}^{2} \right) {{\rm e}^{-i{b}^{2}}}\,. $$
Also, you can get the esult in terms of the incomplete gamma function
$$ {{\rm e}^{-i{b}^{2}}}{\it \Gamma} \left( 0,-i{b}^{2} \right)\,.  $$ 
A: The integral can not be evaluated by the residue theorem, because the contour can not be completed to a closed one. The integral is non-elementary. 
Since $b$ is assumed positive.
$$
   \int_0^\infty \frac{\exp(i b x)}{x+b} \mathrm{d} x = \mathrm{e}^{i b^2} \int_b^\infty \frac{\exp(i b x)}{x} \mathrm{d} x \stackrel{b x = t}{=} \mathrm{e}^{i b^2} \int_{b^2}^\infty \frac{\exp(i t)}{t} \mathrm{d} t = \mathrm{e}^{i b^2} \left( \int_{b^2}^\infty \frac{\cos( t)}{t} \mathrm{d} t + i \int_{b^2}^\infty \frac{\sin(t)}{t} \mathrm{d} t \right)
$$
The latter integrals are known as cosine integral $\operatorname{ci}(b^2)$ and sine integral $\operatorname{si}(b^2)$. 
Thus the integral becomes:
$$
 \int_0^\infty \frac{\exp(i b x)}{x+b} \mathrm{d} x = \left( \operatorname{ci}(b^2) \cos(b^2) - \operatorname{si}(b^2) \sin(b^2) \right) + i \left( \operatorname{si}(b^2) \cos(b^2) + \operatorname{ci}(b^2) \sin(b^2) \right)
$$
