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?
Evaluating $\int\limits_{-\infty}^\infty {\exp(iax)\over1+ix}dx$
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0Yes, this can be done using Fourier Transforms. $\frac{1}{1+ix}$ is the inverse Fourier Transform of$a$simple function of $\omega$ – 2012-10-15
2 Answers
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 $\mathcal{I}=\frac{1}{i}\int^{i\infty}_{-i\infty}\frac{e^{az}}{1+z}\,dz\Rightarrow \mathcal{I}=2\pi\cdot e^{-a}$
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0You suffer from the same problem here as in your answer here http://math.stackexchange.com/questions/1046207/evaluating-an-integral-by-residue-theorem/1046390#1046390 I think you need to understand how inverse Laplace transforms work before publishing "solutions" like these. – 2014-12-01
$\int_{-\infty}^{+\infty}\frac{e^{iax}}{1+ix}dx=\frac{2\pi}{e^a}\quad,\quad a\in\mathbb{R}_+^*$