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

Question

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?

Answer 1

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 $$\mathcal{I}=\frac{1}{i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz\Rightarrow \mathcal{I}=2\pi\cdot e^{-a}$$

Answer 2

$$\int_{-\infty}^{+\infty}\frac{e^{iax}}{1+ix}dx=\frac{2\pi}{e^a}\quad,\quad a\in\mathbb{R}_+^*$$