# Evaluating $\int\limits_{-\infty}^\infty {\exp(iax)\over1+ix}dx$

How does one evaluate the integral $\int\limits_{-\infty}^\infty {\exp(iax)\over1+ix}dx$? I tried Wolfram Alpha, but it just says "computation timed out"... I tried the indefinite integral and got an answer involving some weird function $E_1$. Is it possible to bypass the weird function? I presume the limits of my integral would eliminate that, but how?

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Nothing special about $a$? – J. M. Nov 15 '11 at 0:06
On the other hand, residues look to be the best route for this... – J. M. Nov 15 '11 at 0:08
J.M. is correct. Residue theory is the way to go. However, if you type "exp(iax)/(1+ix)" into Alpha it will give you an alternate form in terms of sines and cosines. If you integrate each term separately (clicking on the link for "give alpha more time" when it times out), it'll give you answers. Then piece them back together. – Bill Cook Nov 15 '11 at 0:16
Fourier maybe ? – Dhaivat Pandya Nov 15 '11 at 0:20
@J.M.: Thanks! :) Unfortunately I don't quite know what "residues" are... And yes, $a$ is just some constant. – owen Nov 15 '11 at 0:46

Set $z=ix$ then $dz=idx$. The integral reads as $$\mathcal{I}=\frac{1}{i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz=\frac{2\pi}{2\pi i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz$$ The integral $$\frac{1}{2\pi i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz=\mathcal{L}^{-1}(\frac{1}{1+z})(a)$$ is the Bromwich integral and the right side is the inverse Laplace transform. To evaluate this inverse Laplace transform consider the following integral $$\mathcal{J}=\oint_{\gamma}\frac{e^{az}}{1+z}\,dz$$ where $\gamma$ is a contour consisting of a vertical line on the imaginary axis and a semicircle on the left-half plane. We could partition the contour integral as follows: $$\oint_{\gamma}\frac{e^{az}}{1+z}\,dz=\int^{iT}_{-iT}\frac{e^{az}}{1+z}\,dz+\oint_{\Gamma_R}\frac{e^{az}}{1+z}\,dz$$ The last integral can be estimated as $$\Big|\oint_{\Gamma_R}\frac{e^{az}}{1+z}\,dz\Big|\leq\oint_{\Gamma_R}\Big|\frac{e^{az}}{1+z}\,dz\Big|\leq \frac{e^{-Ra}}{R-1}\to0$$ as $R\to\infty$. Therefore in the limit the only contribution comes from the first integral namely $$\lim_{T\to\infty}\int^{iT}_{-iT}\frac{e^{az}}{1+z}\,dz=\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz$$ Within this contour there is only one simple pole of the integrand at $z=-1$ with residue $e^{-a}$. Appealing to Cauchy Theorem on residues then $$\oint_{\gamma}\frac{e^{az}}{1+z}\,dz=2\pi i\cdot e^{-a}\Rightarrow \int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz=2\pi i \cdot e^{-a}$$ But $$\mathcal{I}=\frac{1}{i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz\Rightarrow \mathcal{I}=2\pi\cdot e^{-a}$$
$$\int_{-\infty}^{+\infty}\frac{e^{iax}}{1+ix}dx=\frac{2\pi}{e^a}\quad,\quad a\in\mathbb{R}_+^*$$