The goal of this exercise is to determine the Fourier transform of the function𝑓(𝑥)={𝑥if −1<𝑥<10otherwisef(x)={ x0​ if −1<x<1otherwise​

The Fourier transform of a function f(x) is given by the integral:

F(k) = ∫ f(x) e^(-2πikx) dx

In this case, the function f(x) is x for -1 < x < 1 and 0 otherwise. So, we only need to consider the interval from -1 to 1 in the integral. Therefore, the Fourier transform is:

F(k) = ∫_{-1}^{1} x e^(-2πikx) dx

This is an integral that can be solved by integration by parts, where we let u = x and dv = e^(-2πikx) dx. Then du = dx and v = -1/(2πik) e^(-2πikx).

Using the integration by parts formula ∫ u dv = uv - ∫ v du, we get:

F(k) = [-x/(2πik) e^(-2πikx)]{-1}^{1} - ∫{-1}^{1} -1/(2πik) e^(-2πikx) dx

Solving this gives:

F(k) = -1/(2πik) [e^(-2πik) - e^(2πik)] + 1/(4π^2k^2) [e^(-2πik) - e^(2πik)]

Simplifying this using Euler's formula e^(ix) = cos(x) + i sin(x) gives the final result:

F(k) = -1/(πk) sin(2πk) + 2/(π^2k^2) [1 - cos(2πk)]