### Author Topic: Problem 2  (Read 19082 times)

#### Calvin Arnott

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##### Problem 2
« on: November 03, 2012, 03:36:12 PM »
To be perfectly clear- in problem 2 and problem 3, we have that $\beta > 0$, correct?

#### Victor Ivrii

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##### Re: Problem 2
« Reply #1 on: November 03, 2012, 04:40:04 PM »
To be perfectly clear- in problem 2 and problem 3, we have that $\beta > 0$, correct?

Sure you can assume this: $\beta=0$ is covered by (a), (b) and $\beta<0$ is covered by $\beta>0$. However this assumption is not needed.

Cool avatar

#### Hanqing Liu

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##### Re: Problem 2
« Reply #2 on: November 06, 2012, 06:11:27 PM »
I think I'm on the wrong track or this question just doesn't seem to be integrable!
« Last Edit: November 06, 2012, 07:08:13 PM by Hanqing Liu »

#### Victor Ivrii

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##### Re: Problem 2
« Reply #3 on: November 06, 2012, 07:25:14 PM »
I think I'm on the wrong track or this question just doesn't seem to be integrable!

Read preamble! Use properties!!! See lectures

#### Zarak Mahmud

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##### Re: Problem 2
« Reply #4 on: November 07, 2012, 05:26:22 PM »
This integral can be computed using residues, but I think the whole point of the Fourier tranform is lost if one uses that approach...

#### Aida Razi

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##### Re: Problem 2
« Reply #5 on: November 07, 2012, 09:30:35 PM »
Solution is attached!

#### Ian Kivlichan

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##### Re: Problem 2
« Reply #6 on: November 07, 2012, 09:35:05 PM »
Aida: How did you do the integral?

#### Calvin Arnott

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##### Re: Problem 2
« Reply #7 on: November 07, 2012, 09:36:22 PM »
Problem 2

Let: $\alpha > 0, \beta > 0, n\in\mathbb{N}$. Compute the Fourier transform for:

a. $\left(x^2 + \alpha^2\right)^{-1}$

Let $f\left(z\right)$ be holomorphic on a domain $D$. Cauchy's formula states that for any positively oriented piecewise smooth simple closed curve $\gamma \in D$ whose inside $\Omega$ lies in $D$:

$$\forall z \in \Omega: f\left(z\right) = \frac{1}{2 \pi i}\int_\gamma \frac{f\left(\zeta\right)}{\left(\zeta - z\right)} d\zeta$$

$$\text{Claim: } F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{\left(x - i\alpha\right)\left(x + i\alpha\right)}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha }$$

Proof: Let: $\gamma_r^+$ be the positively oriented semi-circle on the upper half-plane with radius $r$, and $\gamma_r^-$ be the positively oriented semi-circle on the lower half-plane with radius $r$. Then both $\{\gamma_r^+,\gamma_r^-\}$ are positively oriented piecewise smooth simple closed curves. Moreover, $f^+\left(z\right)= \frac{e^{-i k z}}{\left(z + i\alpha\right)}$ and $f^-\left(z\right) = \frac{e^{-i k z}}{\left(z - i\alpha\right)}$ are holomorphic on $\gamma_r^+$ and $\gamma_r^-$ respectively, as $e^{z}$ is entire and we avoid the poles of the rational functions on their respective domains. We take cases and apply Cauchy's formula:

$$\text{Suppose: } k < 0. \text{ Then: } \int_{\gamma_r^+}\frac{\frac{e^{-i k x}}{\left(x + i\alpha\right)}}{\left(x - i\alpha\right)}dx = 2 \pi i f^+\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(i \alpha\right) + i\alpha\right)} = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\text{Suppose: } k > 0. \text{ Then: } \int_{\gamma_r^-}\frac{\frac{e^{-i k x}}{\left(x - i\alpha\right)}}{\left(x + i\alpha\right)}dx = 2 \pi i f^-\left(i\alpha\right) = 2\pi i \frac{e^{-i k \left(i \alpha\right)}}{\left(\left(-i \alpha\right) - i\alpha\right)} = -\frac{ \pi e^{- \alpha k} }{\alpha }$$

$$\text{Now, } \int_{\gamma_r}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \text{ is the sum of two integrals:}$$

$$\int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{-r}^{r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\int_{\gamma_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = \int_{r}^{-r}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + \int_{C_r^-}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx = -\frac{ \pi e^{- \alpha k} }{\alpha }$$

Where $C_r$ is the upper or lower half of the half circle. We show that the $C_r$ portion of the integral vanishes as $r \rightarrow \infty$. Without loss of generality we set $\alpha$ to $1$ to simplify the inequalities:

$$\mid \int_{C_r^+}\frac{e^{-i k x}}{x^2 + 1}dx \mid \le \int_{C_r^+} \mid \frac{e^{-i k x}}{x^2 + 1} \mid \mid dx \mid \le \int_{C_r^+}\frac{1}{r^2 - 1} \mid dx \mid = \frac{\pi r}{r^2 - 1} \rightarrow 0 \text{ as } r \rightarrow \infty$$

$$\text{Then: } \lim_{r \to +\infty} \int_{\gamma_r^+}\frac{e^{-i k x}}{\left(x + i\alpha\right)\left(x - i\alpha\right)}dx \rightarrow \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx + 0 = \frac{ \pi e^{\alpha k} }{\alpha }$$

$$\implies \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{\alpha k} }{\alpha } \text{ when } k < 0$$

And similarly for our $\gamma_r^-$ integral, changing the bounds of integration and the sign. We have then: $\frac{ \pi e^{\alpha k} }{\alpha }$ for $k < 0$ and $\frac{ \pi e^{-\alpha k} }{\alpha }$ for $k > 0$. Thus:

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \blacksquare$$

b. $x\left(x^2 + \alpha^2\right)^{-1}$

We have for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$. Then:
$$g\left(x\right) = x\left(x^2 + \alpha^2\right)^{-1} = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i \partial_k \left(\frac{ \pi e^{-\alpha |k|} }{\alpha }\right) = - i \pi sgn\left(k\right) e^{-\alpha |k|} \blacksquare$$

Where $f\left(x\right) = \left(x^2 + \alpha^2\right)^{-1}$ and $F\left(k\right) = \frac{ \pi e^{-\alpha |k|} }{\alpha }$ as found in part a.

c. i) $\left(x^2 + \alpha^2\right)^{-1} \cos\left(\beta x\right)$

Two properties of the Fourier transform are that for: $g\left(x\right) = e^{i a x} f\left(x\right)$ the Fourier transform of $g\left(x\right)$ is given by $G\left(k\right) = F\left(k-a\right)$, and that the transform is linear: $h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right)$

$$\text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$\left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.}$$

$$\implies G\left(k\right) = \frac{1}{2} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } + \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare$$

ii) $\left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$

$$\text{Proceeding as in part c. i): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:}$$

$$\left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{e^{-i k x}}{x^2 + \alpha^2}dx = \frac{ \pi e^{-\alpha |k|} }{\alpha } \text{ is our transform from part a.}$$

$$\implies G\left(k\right) = \frac{1}{2i} \left( \frac{ \pi e^{-\alpha |k - \beta|} }{\alpha } - \frac{ \pi e^{-\alpha |k + \beta|} }{\alpha }\right) \blacksquare$$

d. i) $x \left(x^2 + \alpha^2\right)^{-1} \cos\left(\beta x\right)$

$$\text{Proceeding as in part c. i): } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$x \left(x^2 + \alpha^2\right)^{-1}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} + e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.}$$

$$\implies G\left(k\right) = \frac{1}{2} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right)$$

$$= - \frac{i \pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} + sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare$$

ii) $x \left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$

$$\text{Proceeding as in part c. ii): } \sin{\beta x} = \frac{1}{2i}\left(e^{i \beta x} - e^{- i \beta x}\right) \text{ so we write:}$$

$$x \left(x^2 + \alpha^2\right)^{-1}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x \left(x^2 + \alpha^2\right)^{-1} - e^{- i \beta x} x \left(x^2 + \alpha^2\right)^{-1}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = \int_{-\infty}^{\infty}\frac{x e^{-i k x}}{x^2 + \alpha^2}dx = - i \pi sgn\left(k\right) e^{-\alpha |k|} \text{ is our transform from part b.}$$

$$\implies G\left(k\right) = \frac{1}{2i} \left( - i \pi sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} {\alpha } + i \pi sgn\left(k + \beta\right) e^{-\alpha |k + \beta|} {\alpha }\right)$$

$$= - \frac{\pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare$$
« Last Edit: November 18, 2012, 04:56:09 PM by Calvin Arnott »

#### Calvin Arnott

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##### Re: Problem 2
« Reply #8 on: November 07, 2012, 09:36:34 PM »
Part 2
« Last Edit: November 07, 2012, 11:50:15 PM by Calvin Arnott »

#### Zarak Mahmud

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##### Re: Problem 2
« Reply #9 on: November 07, 2012, 09:39:25 PM »
Aida: How did you do the integral?

She didn't. She used a fourier transform pair from part 1(a).
« Last Edit: November 08, 2012, 12:15:14 AM by Zarak Mahmud »

#### Aida Razi

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##### Re: Problem 2
« Reply #10 on: November 07, 2012, 09:49:47 PM »
Aida: How did you do the integral?

He didn't. He used a fourier transform pair from part 1(a).

Yes, That's right!
By the way Zarak, I guess it is clear from my name that I am female!

#### Victor Ivrii

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##### Re: Problem 2
« Reply #11 on: November 07, 2012, 09:53:03 PM »
This integral can be computed using residues, but I think the whole point of the Fourier tranform is lost if one uses that approach...

Yes, residues work miracles here (see Calvin) but most have not taken Complex Analysis and there is a property for that

$\newcommand{\sgn}{\operatorname{sgn}}$
Calvin, it is $\sgn$ not $sgn$

See source
« Last Edit: November 07, 2012, 09:57:03 PM by Victor Ivrii »

#### Ziting Zhou

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##### Re: Problem 2
« Reply #12 on: November 07, 2012, 10:00:44 PM »
Hi. I answered 2(d) in four cases. Is it not necessary to do it?

#### Victor Ivrii

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##### Re: Problem 2
« Reply #13 on: November 07, 2012, 10:07:01 PM »
Ziting, your jpgs are borked. Anyway, we have enough solutions.

Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

#### Ziting Zhou

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##### Re: Problem 2
« Reply #14 on: November 07, 2012, 10:16:34 PM »
Ziting, your jpgs are borked. Anyway, we have enough solutions.

Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

Yes, there are enough solutions. So I won't post mine. I am still wondering if we need to consider the problem on cases, for example, the transform is not defined when w=Î² or w=-Î². Thanks!