Answer:
the Fourier transform of 1/(1+x^2) is F(k) = sqrt(pi/2) * (e^(-k) - e^(k)).
Explanation:
The Fourier transform of a function f(x) is defined as:
F(k) = (1/sqrt(2*pi)) * integral from -infinity to infinity of f(x) * e^(-ikx) dx
In this case, we want to find the Fourier transform of f(x) = 1/(1+x^2):
F(k) = (1/sqrt(2*pi)) * integral from -infinity to infinity of 1/(1+x^2) * e^(-ikx) dx
We can evaluate this integral using complex analysis techniques. One way is to use the residue theorem and evaluate the residues of the integrand at the poles of 1/(1+x^2). The poles of this function are at x = i and x = -i.
Using the residue theorem, we can write:
F(k) = 2pii * Res(f(x)e^(-ikx), x=i) + 2pi*i * Res(f(x)*e^(-ikx), x=-i)
The residues at x = i and x = -i are:
Res(f(x)*e^(-ikx), x=i) = lim(x->i) [(x-i)*f(x)*e^(-ikx)]
= 1/(2i) * e^(-k)
Res(f(x)*e^(-ikx), x=-i) = lim(x->-i) [(x+i)*f(x)*e^(-ikx)]
= -1/(2i) * e^(k)
Substituting these into the formula for F(k), we get:
F(k) = pi * (e^(-k) - e^(k))/(sqrt(2*pi))
Simplifying this expression, we get:
F(k) = sqrt(pi/2) * (e^(-k) - e^(k))