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Topics - Tristan Fraser

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Quiz-5 / Quiz 5, T5102
« on: March 07, 2018, 06:28:31 PM »
Let $\a > 0 $ , find the fourier transforms of $$ (x^2 + a^2)^{-1} $$ and $$ x(x^2 + a^2)^{-1} $$. HINT: Consider the fourier transform of $e^{-\beta|k|} $.

Starting with the hint, we can take:

$ f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ikx} e^{-\beta|k|} dk = \frac{1}{\sqrt{2\pi}}[ \int_{0}^{\infty} e^{-k(\beta - ix)} dk + \int_{-\infty}^{0} e^{k(\beta + ix)} dk] $

Integrating and evaluating at the bounds will leave us:

$ \frac{1}{\sqrt{2\pi}} [\frac{1}{\beta + ix} + \frac{1}{\beta - ix} ] = \frac{2\beta}{x^2 + \beta^2} \times \frac{1}{\sqrt{2\pi}} $

Then, for problem 1:

let $\beta = a $

Then using the property that $\hat{f} = F $ and $\hat{F} = f $, we can note:

the fourier transform of the above is merely $\frac{1}{x^2 + a^2} \ times \frac{2a}{\sqrt{2\pi}} $. The above property then implies that some multiple of the function $e^{-a|k|}$ is our desired function for the fourier transform, specifically:

$g(k) = \frac{\sqrt{2\pi}}{2a} e^{-a|k|}$ 's fourier transform, $\hat{g(x)} = \frac{\sqrt{2\pi}}{2a} \ times \frac{2\beta}{x^2 + a^2} \times \frac{1}{\sqrt{2\pi}}  = \frac{1}{x^2 + a^2}$ as desired.

For the second function's fourier transform: note that $g(x) = xf(x) \rightarrow \hat{g(k)} =  i\hat{f'(k)} $ is a property, and that the second function is x times the previous problem's function.

Thus, take the derivative of $\hat{f(k)} $ i.e. $\frac{d}{dk} ( e^{-a|k|}) =  \frac{-ak e^{-a|k|}}{|k|} $ So then:

the forward fourier transform is:

$\frac{-i\sqrt{\pi}k e^{-a|k|}}{\sqrt{2}|k|}$







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