Proving inverse Fourier transform of 1/(1+x^2) = 1/(1+x^2)

Vitani11
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Homework Statement


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
In other words the question asks to solve this integral: 1/sqrt(2π) ∫F(t)eitxdt and show that it equals 1/(1+x2)

Homework Equations


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
1/sqrt(2π) ∫F(t)eitxdt

The Attempt at a Solution


I want to use complex numbers but these functions are analytic and so I can not do this using residues. A naive approach to integrating with respect to t is straightforward but the limits cause the result to be infinity. How should I approach this properly?
 
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Vitani11 said:

Homework Statement


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
In other words the question asks to solve this integral: 1/sqrt(2π) ∫F(t)eitxdt and show that it equals 1/(1+x2)

Homework Equations


F(t) = sqrt(π/2)e-t for t>0
F(t) = sqrt(π/2)et for t<0
1/sqrt(2π) ∫F(t)eitxdt

The Attempt at a Solution


I want to use complex numbers but these functions are analytic and so I can not do this using residues. A naive approach to integrating with respect to t is straightforward but the limits cause the result to be infinity. How should I approach this properly?
Break up the real line into ##-(\infty, 0)## and ##(0,\infty)##, then do two separate integrals. Each integral is really easy.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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