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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.

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It seems there is no elementary result possible. See wolframalpha.com/input/?i=\int^{\infty}_{0}\frac{e^{ix}}{x%2Bb} –  Bombyx mori Aug 24 '12 at 11:24
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Since $0$ has nothing special to do with the integrand, it is unlikely that contour methods will help. –  GEdgar Aug 24 '12 at 13:02

2 Answers 2

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

$$ {{\rm e}^{-i{b}^{2}}}{\it \Gamma} \left( 0,-i{b}^{2} \right)\,. $$

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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: $$ \int_0^\infty \frac{\exp(i b x)}{x+b} \mathrm{d} x = \left( \operatorname{ci}(b^2) \cos(b^2) - \operatorname{si}(b^2) \sin(b^2) \right) + i \left( \operatorname{si}(b^2) \cos(b^2) + \operatorname{ci}(b^2) \sin(b^2) \right) $$

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