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What is the Fourier Cosine Transform of 1/(1+x^2) ?
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This problem is a good example of how to use contour integration and various theorems on Fourier Transforms to calculate the integral (which makes it worthwhile even though this looks like a homework problem).

First note that f(x) = 1/(1+x^2) is an even function, which means it has the same sign as x goes from positive to negative values. This means that its cosine transform is equal to its full Fourier Transform (FT), since the odd portion (sine transform) cancels out. So we can write

FT[f(x)] = Integral{ e^-2πixs/(1+x^2) }  -∞ < x < ∞   <--  usual definition of FT with s = transform variable and where

e^-2πixs = cos(2πixs) + isin(2πixs).

This is an integral along the real axis. It is not particularly obvious how to perform the integration using the usual algebraic manipulations. However, a standard approach at this point is to solve it along a contour in the complex plane (which includes the real axis).

To this end, first note that

1/(x^2+1) = 1/[ (x-i)(x+i) ]   <-- i = √(-1)

which means there are 2 singularities at x = i and x = -i, neither of which are on the real axis (!?). Although it may seem like we haven't made too much progress, we can use some powerful results from complex analysis to evaluate the integral, namely that the value of the contour integral is related to the value of the function at a singularity inside the contour.

Let x -> z = complex variable and define a contour in the complex plane consisting of the real axis from -R < z < R and connected at both ends by a semi-circle in the upper half plane from R counter-clockwise to -R; i.e., along z=Re^iθ for 0 ≤ θ ≤ π. The contour is then expanded by letting R -> ∞. This contour encloses the singularity at z = i, which is the whole point of all these manipulations

The integral can be split into the piece along the x axis, from -∞ to ∞ (which is the integral we want), and the piece along the semi-circle stretching out to infinity. Because 1/(z^2+1) goes to zero as |z| -> ∞ and that e^(-i2πzs) ≤ 1 for all z, the integral along the infinite semi-circle goes to zero. Thus, the value of the integral along the whole contour, which can be calculated using Residue Theory (below), is equal to its value along the real axis, which is the integral we want.

So now we can evaluate the integral. Without going into detail, I'm just going to have to state that, from Residue Theory, the integral is equal to

2πi・{(z-i)f(z)} evaluated at z = i. Plugging in using the form of f(z) gives (notice that the z-i term cancels)

2πi{ [(z-i)e^(-i2πzs]/[(z-i)(z+i)] at z = i --> 2πi・e^(-2πs)/2i = πe^(-2πs), which is your answer.

I hope you have had some experience with complex analysis so that this doesn't seem too mysterious. There are probably ways to use real analysis to obtain the result, the the method shown here is really the one that jumps out as the best and easiest way.  

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