Answer to Question #102649 in Calculus for BIVEK SAH

Question

Determine the Fourier transform of the function

f(t)={sin3t -pie<=t<=+pie

=0 otherwise

Answer 1

$F_t[f]=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}f(t)e^{-iwt}dt=\frac{1}{\sqrt{2\pi}}\int\limits_{-\pi e}^{\pi e}sin3te^{-iwt}dt=$

$=\frac{1}{\sqrt{2\pi}}\frac{e^{-iwt}3\cos3t-iw\sin3t}{w^2-9}_{-\pi e}^{\pi e}=\frac{i}{\sqrt{2\pi}}\left( \frac{\sin\pi e(w+3)}{w+3}-\frac{\sin\pi e(w-3)}{w-3}\right)$