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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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