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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 $a$?2011-11-15
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    On the other hand, residues look to be the best route for this...2011-11-15
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    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.2011-11-15
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    Fourier maybe ?2011-11-15
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    @J.M.: Thanks! :) Unfortunately I don't quite know what "residues" are... And yes, $a$ is just some constant.2011-11-15
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    Okay... how good is your complex analysis?2011-11-15
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    @J.M.: Erm, I'm a beginner...?:S but I'm happy to learn2011-11-15
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    @BillCook: Ah, I've never noticed that "button"!2011-11-15
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    I know $a$ is a constant; I was wondering if you were assuming, for instance, that $a$ is real. (If $a$ is purely imaginary, then we've no finite value to speak of.)2011-11-15
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    @J.M.: Sorry about that, didn't mean to insult you! :S Anyway, yes, $a\in \mathbb R$2011-11-15
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    No worries, I never read it as an insult. :) I was actually trying to help you make your question more precise, complex analysis being tricky and all. What works nicely on the real line can get weird in the complex plane, you see...2011-11-15
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    In any event, any good book on complex analysis should have a chapter/section on residues.2011-11-15
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    @J.M.: Thanks! :) I'll try to get hold of one...2011-11-15
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    Can you do the case $a=0$?2011-11-15
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    for $a = 1$ we get $\dfrac {2\pi} {e}$2012-04-26
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    This looks like the _Fourier Transform_.2012-07-05
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    Yes, 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 2

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

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    This is not quite correct. Think about the integral for the inverse transform.2014-06-12