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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 $a$? –  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
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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. –  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

1 Answer 1

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

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This is not quite correct. Think about the integral for the inverse transform. –  Ron Gordon Jun 12 at 12:15

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