Another Fourier Inversion Problem

[cloud18]

Find the inverse transform of

$$\frac{\sin{a\omega}}{\omega}$$

I know the step function has a transform of this form, so I as able to find the inverse transform by assuming it was some step function and then looked for the right constants.

However, I would like to also know how to do it by the definition:

$$f(x) = \frac{1}{2\pi} \lim{\int{\overline{f}(\omega)e^{-i\omega x} d\omega}}$$

Where the limit is L--> infinity and the integration limits are -L to +L.

I think this must be done by contour integration? Can someone show me how to setup the contour integral?

 

[lippyka]

Thanks.The inverse transform of $\frac{\sin{a\omega}}{\omega}$ is given by:f(x) = \frac{1}{2\pi} \lim_{L\rightarrow \infty}\int_{-L}^{L} \frac{\sin{a\omega}}{\omega} e^{-i\omega x} d\omega To do this using contour integration, we set up a semicircle in the upper half plane with radius L and then we need to break up the integral into two parts: one part on the semicircle and one part on the real line. The integral over the semicircle will go to zero as L goes to infinity. The integral on the real line is just the original integral we wanted to solve.We have:f(x) = \frac{1}{2\pi} \lim_{L\rightarrow \infty}\left[ \int_{-L}^{L} \frac{\sin{a\omega}}{\omega} e^{-i\omega x} d\omega + \int_{C_L} \frac{\sin{a\omega}}{\omega} e^{-i\omega x} d\omega \right]Where $C_L$ is the semicircle in the upper half plane with radius L. Since the integrand is analytic on the entire region enclosed by the contour, and since the integral along the arc of the semicircle goes to zero as $L \rightarrow \infty$, the integral over the semicircle can be ignored. So, we have:f(x) = \frac{1}{2\pi} \lim_{L\rightarrow \infty}\int_{-L}^{L} \frac{\sin{a\omega}}{\omega} e^{-i\omega x} d\omega Which is the same as the original integral we wanted to solve.