Author Topic: 5.2P, Question 3 approach  (Read 1029 times)

Tristan Fraser

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5.2P, Question 3 approach
« on: March 06, 2018, 09:59:05 PM »
For 3a we are asked to take the Fourier transform of

$$(x^2 + a^2) ^ {-1}$$ For $a> 0$

i.e. to find $\tilde{f(k)} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} f(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} (x^2 + a^2) ^ {-1} dx$ and using Euler's identity, we can decompose the integral into cosines and sines (unlike in problems 2, where the trick seemed to lie doing the reverse) giving us:

$$\tilde{f(k)} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) (coskx - isinkx) \frac{1}{x^2 + a^2} dx$$

The denominator is even, the first term is even , so we'll keep it (as it is even), the second term is odd, with an even denominator. Integrating over all space means that the integral of any odd function will be 0, leaving only the even function integral. Furthermore, since even functions are symmetric about x = 0, we can rewrite the integral:

$$\tilde{f(k)} = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty}\frac{coskx}{x^2 + a^2} dx$$

From here, I'm not quite sure how to approach this integral, as I'd normally try a trig substitution, but it seems like it would only complicate this...

Am I missing something obvious? What steps lack justification/ are incorrect?

NOTE: This question also applies to the approach in general for these problems in 3, as 3b,3c, and 3d are complications on this/add on more steps with integration by parts, so I'd also like to know if this even is the right treatment for this class of problems

Victor Ivrii

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Re: 5.2P, Question 3 approach
« Reply #1 on: March 07, 2018, 03:13:34 AM »
There are two different ways to deal with it:

1) Before that we found F.T. of $e^{-b|x|}$; what we got? And F.T. and I.F.T. are rather interchangable

2) If you studied complex variables ...

Obseve that $e^{-ikz}$ is bounded for real $k$ and $\renewcommand{\Im}{\operatorname{Im}}\Im z <0$. Then
$(2\pi)^{-1}\int _\Gamma e^{-ikz}(z^2+a^2)^{-1}\,dz$ equals to the sum of residues of the integrand (multiplied by $i$) inside of $\Gamma$, where $\Gamma$ is a simple counter-clock-wise contour. We take it : from $(R,0)$ to $(-R,0)$ along real axis, and then back along half-circle in the lower complex half-plane. When $R\to \infty$ the second integral tends to $0$ (easy to estimate) and the first to what we need, albeit with a sign $-$.

There is just a simple pole $-ai$ there
« Last Edit: March 07, 2018, 07:57:34 AM by Victor Ivrii »

Jingxuan Zhang

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Re: 5.2P, Question 3 approach
« Reply #2 on: March 07, 2018, 06:54:30 AM »
If you used Fisher's text in MAT344, you can find it on p. 344. Factor the denominator and use CIF/Residual theorem provided a proper orientation is taken to make sure the integral vanish on the arc of the "semicircle contour".

The following questions are not meant to be dealt starting from scratch. Referring to the properties list in section 5.2 I found all them follow in one way or another the first question, the "base" case.
« Last Edit: March 07, 2018, 06:56:15 AM by Jingxuan Zhang »

Tristan Fraser

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Re: 5.2P, Question 3 approach
« Reply #3 on: March 07, 2018, 12:00:46 PM »
So for 3a, we know that the fourier transform of $f(x) = e^{-|x|\alpha}$ is (I've shortened several steps)
$$\tilde{f}(k) = \frac{1}{\sqrt{2\pi}} ( \int_{0}^{\infty} e^{-(\alpha + ik)x} dx + \int_{-\infty}^{0} e^{x(\alpha - ik)} dx ) = \frac{1}{\sqrt{2\pi}} (\frac{1}{\alpha + ik} - \frac{1}{\alpha - ik} = \frac{\sqrt{2}\alpha}{\sqrt{\pi}(\alpha^2 + k^2)} = \tilde{f(k)}$$

Then use the property of $\tilde{F} = f$ and $\tilde{f} = F$, where the above result is $\tilde{F}(x)$, thus, by the property, the fourier transform of the function $f(x) = \frac{1}{x^2 + \alpha^2}$. Then the Fourier transform is just $\tilde{f}(k) =\frac{\sqrt{2\pi}}{2\alpha} e^{-|\alpha|k}$ ?

Then for the other parts, this can just be used for integrations by parts as a substituted value?

« Last Edit: March 07, 2018, 01:48:47 PM by Victor Ivrii »

Victor Ivrii

PS. We use hat'' rather than tilde'' and we use a bit different numerical factors. As long as you are consistent, it is fine