### Author Topic: HA7-P5  (Read 1074 times)

#### Victor Ivrii

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##### HA7-P5
« on: November 01, 2015, 05:10:24 PM »

#### Xi Yue Wang

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##### Re: HA7-P5
« Reply #1 on: November 03, 2015, 08:11:49 PM »
For part (a), given that $|x| \leq a$, we get $$\hat{f}(\omega) = \frac{1}{2\pi}\int_{-a}^{a} e^{-i\omega x } dx\\= \frac{1}{2\pi}\frac {e^{-i\omega a}-e^{i\omega a}}{(-i\omega)}\\=\frac{\sin(\omega a)}{\pi\omega}$$

For part (b), we take $g(x) = xf(x)$, $\hat{g}(\omega) = i\hat{f'}(\omega)$, where $f(x)$ is defined in part (a).$$\hat{f'}(\omega) = \frac{a\cos(\omega a)}{\pi\omega} - \frac{\sin(\omega a)}{\pi\omega^2}\\\hat{g}(\omega) = i( \frac{a\cos(\omega a)}{\pi\omega} - \frac{\sin(\omega a)}{\pi\omega^2})$$

For part (c), by Inverse Fourier Transformation, $$\int_{-\infty}^{\infty} \hat{f}(\omega)e^{i\omega x} d\omega=\int_{-\infty}^{\infty} \frac{\sin(\omega a)}{\pi\omega}e^{i\omega x} d\omega= f(x) = 1$$ where $|x|\leq a$
We take $x=0, a=1$, then we get $$\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega = 1$$
Hence, $$\int_{-\infty}^{\infty} \frac{\sin(\omega)}{\omega} d\omega = \pi$$