Question

Determine the Fourier transform of the function
F(t) = {1 - t , 0 ≤ t ≤ 1
= {1 + t , -1≤ t ≤ 0
= {0 , otherwise

Answer 1

From the definition of the Fourier transformation

$$f(x)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}F(t)e^{-itx}dt$$

we obtain

$$f(x)=\int\limits_{-1}^{0}(1+t)e^{-itx}dt+\int\limits_{0}^{1}(1-t)e^{-itx}dt=$$ $$=2\int\limits_{0}^{1}\cos(tx)dt-2\int\limits_{0}^{1}t\cos(tx)dt=2\frac{\sin(x)}{x}+2\frac{1}{x^2}-2\frac{\cos (x)}{x^2}-2\frac{\sin (x)}{x}=$$ $$=2\frac{1-\cos(x)}{x^2}$$

Answer:

$$f(x)=2\frac{1-\cos(x)}{x^2}$$