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I am seeking the method for calculating the following integral

$$\int_{-\infty}^\infty\frac{e^{-2ix\pi\psi}}{1+x^2} dx $$

Ideas I have are:

1) substition (however which one?) 2) integration by parts

The integral comes from the Fourier transform of $$\frac{1}{1+x^2}$$

Cameron Buie
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unseen_rider
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  • What is $\psi$? – Cameron Buie Jul 06 '16 at 11:42
  • $$\psi$$ comes from the Fourier transform definition - $$F(\psi)=$$ integral in question. – unseen_rider Jul 06 '16 at 11:46
  • The integral can be calculated easily with complex methods. – Gennaro Marco Devincenzis Jul 06 '16 at 12:06
  • @gennaromarcodevincenzis like contour integration? – unseen_rider Jul 06 '16 at 12:40
  • Yes, and if you search for it on the site there's a high probability it's been done. The function has simple poles at $\pm i$ and it can be done with a semicircular contour, showing that the "upper" part of the integral vanishes for $R \rightarrow \infty$, where $R$ is the radius of the semicircle.

    If you don't like complex methods (but you should like them, in this case I think is by far the easiest method) I'm fairly certain that cleverly inserting a parameter and differentiating under the integral sign works just as well.

     – Gennaro Marco Devincenzis Jul 06 '16 at 12:43

1 Answers1

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$$\int_{-\infty}^\infty\frac{e^{-2ix\pi\psi}}{1+x^2} dx =\int_{-\infty}^\infty\frac{\cos(2\pi\psi\,x)}{1+x^2}dx-i\int_{-\infty}^\infty\frac{\sin(2\pi\psi\,x)}{1+x^2}dx$$ Please check this question

$$\color{red}{I(\lambda)=\int_{-\infty}^{\infty}{\cos(\lambda x)\over x^2+1}dx=\frac{\pi}{e^{\lambda}}}$$ and

$$\color{red}{J(\lambda)=\int_{-\infty}^{\infty}{\sin(\lambda x)\over x^2+1}dx=0}$$

Now set $\lambda=2\pi\psi$

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Behrouz Maleki
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