Method for calculating integral of $e^{-2ix\pi\psi}/(1+x^2)$

514 Views Asked by At

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 $$\frac{1}{1+x^2}$$

1

There are 1 best solutions below

0
On BEST ANSWER

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