# Toronto Math Forum

## APM346-2018S => APM346--Tests => Quiz-5 => Topic started by: Tristan Fraser on March 07, 2018, 06:28:31 PM

Title: Quiz 5, T5102
Post by: Tristan Fraser 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:

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

Title: Re: Quiz 5, T5102
Post by: George Lu on March 07, 2018, 06:51:46 PM
I went with a more brute force method, wherein I think I made a minor mistake (I did not have |k| at the end like we should have gotten) so it would be great to find out where.

I immediately plugged in the function $f(x)=\frac{1}{(x^2+a^2)}$ into the integral $F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-ikx}}{(x-ia)(x+ia)}$ (where I rewrote the denominator to make use of residue theorem ($\int f = 2\pi i Res (f;-ia)$)), and integrated over the contour:

Giving me $F(k)=\frac{\sqrt{\pi}}{\sqrt{2}a}e^{-ak}$

For the 2nd part, I simply applied the property that $g(x)=x*f(x)\rightarrow G(k)=ikF(x)$
Title: Re: Quiz 5, T5102
Post by: Jingxuan Zhang on March 08, 2018, 07:23:50 AM
George:
The reason that the absolute value is missing is that the contour you choose must depend on the sign of $k$ If $k<0$ then your contour should instead be upper semicircle and if $k>0$ then you are right. This is done to satisfy the hypothesis of Jordan's lemma, which you implicitly used to control the integral over the arc as it increasese.

Also what property do you refer to? I think $\hat{xf}=i\hat{f}'$?

Beside: don't we always use $\frac{1}{2\pi}$ for the scaling? and this for one thing at least better suit Residual Theorem.
\hat{f}(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{(x-ai)(x+ai)}=\left\{\begin{align*}\left.i\frac{e^{-i\omega x}}{x+ai}\right|_{x=ai}&&\omega<0\\ \left.-i\frac{e^{-i\omega x}}{x-ai}\right|_{x=-ai}&&\omega>0\end{align*}\right.=\frac{e^{-|\omega|a}}{2a}
And so immediately
$$\hat{xf}(\omega)=i\hat{f}'(\omega)=-i \text{sgn}(\omega)\frac{e^{-|\omega|a}}{2}$$
Title: Re: Quiz 5, T5102
Post by: Victor Ivrii on March 08, 2018, 07:55:12 AM
Good job (Tristan, George,  Jingxuan)

Now remarks:

\begin{document}\documentclass[tikz]{standalone}\usepackage{mathtools}\renewcommand{\Re}{\operatorname{Re}}\renewcommand{\Im}{\operatorname{Im}}\begin{tikzpicture}\draw [thin,->] (-4,0)--(4,0) node[below]{$\Re z$};\draw [thin,->] (-0,-4)--(0,.5) node[right]{$\Im z$};\draw [thick, red,->] (3,0) arc(0:-180:3);\draw [thick, red,->] (-3,0)--(3,0);\fill[red] (0,-1) circle (0.05) node[left] {$-ai$};\end{tikzpicture}\end{document}