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?
 A: 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 $$\mathcal{I}=\frac{1}{i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz\Rightarrow \mathcal{I}=2\pi\cdot e^{-a}$$
A: $$\int_{-\infty}^{+\infty}\frac{e^{iax}}{1+ix}dx=\frac{2\pi}{e^a}\quad,\quad a\in\mathbb{R}_+^*$$
