# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 6 => Topic started by: Calvin Arnott on November 03, 2012, 03:36:12 PM

Title: Problem 2
Post by: Calvin Arnott on November 03, 2012, 03:36:12 PM
To be perfectly clear- in problem 2 and problem 3, we have that $\beta > 0$, correct?
Title: Re: Problem 2
Post by: Victor Ivrii 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
Title: Re: Problem 2
Post by: Hanqing Liu 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!
Title: Re: Problem 2
Post by: Victor Ivrii 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
Title: Re: Problem 2
Post by: Zarak Mahmud 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...
Title: Re: Problem 2
Post by: Aida Razi on November 07, 2012, 09:30:35 PM
Solution is attached!
Title: Re: Problem 2
Post by: Ian Kivlichan on November 07, 2012, 09:35:05 PM
Aida: How did you do the integral?
Title: Re: Problem 2
Post by: Calvin Arnott 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) $x \left(x^2 + \alpha^2\right)^{-1} \sin\left(\beta x\right)$

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

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

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

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

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

$$= - \frac{\pi}{2} \left(sgn\left(k - \beta\right) e^{-\alpha |k - \beta|} - sgn\left(k + \beta\right) e^{-\alpha |k + \beta|}\right) \blacksquare$$
Title: Re: Problem 2
Post by: Calvin Arnott on November 07, 2012, 09:36:34 PM
Part 2
Title: Re: Problem 2
Post by: Zarak Mahmud 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).
Title: Re: Problem 2
Post by: Aida Razi 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!
Title: Re: Problem 2
Post by: Victor Ivrii 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
Title: Re: Problem 2
Post by: Ziting Zhou on November 07, 2012, 10:00:44 PM
Hi. I answered 2(d) in four cases. Is it not necessary to do it?
Title: Re: Problem 2
Post by: Victor Ivrii 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
Title: Re: Problem 2
Post by: Ziting Zhou 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!
Title: Re: Problem 2
Post by: Fanxun Zeng on November 07, 2012, 10:26:05 PM
I agree with Ziting Zhou that we CAN divide 4 cases in part d if we wish, exactly what I did. Let me show mathematically what the four cases are: 1) w> Î²>0 2) Î² > w > 0 3) w<0 and |w|<|Î²| 4) w<0 and |w|>|Î²|  if we do want to divide. Thanks!
Title: Re: Problem 2
Post by: Calvin Arnott on November 07, 2012, 11:54:42 PM
Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

I changed the .pdfs to png's for now, when I have some time later I'll try to massage the tex code into mathjax/forum format.
Title: Re: Problem 2
Post by: Zarak Mahmud on November 08, 2012, 12:19:45 AM
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!

Hahah oops, sorry!
Title: Re: Problem 2
Post by: Zarak Mahmud on November 08, 2012, 12:21:22 AM
Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

I changed the .pdfs to png's for now, when I have some time later I'll try to massage the tex code into mathjax/forum format.

Calvin - just curious, what program are you in?
Title: Re: Problem 2
Post by: Kun Guo on November 14, 2012, 10:00:01 PM
Same question as Ziting, should there be four cases regarding  beta and +/- omega relationship? The sign will change when we get rid of the absolute sign and take the derivative.
Title: Re: Problem 2
Post by: Victor Ivrii on November 15, 2012, 01:21:59 AM
Same question as Ziting, should there be four cases regarding  beta and +/- omega relationship? The sign will change when we get rid of the absolute sign and take the derivative.

You don't need any divisions, just apply properties. And I already demonstrated http://www.math.toronto.edu/courses/apm346h1/20129/L17.html#sect-17.4 (http://www.math.toronto.edu/courses/apm346h1/20129/L17.html#sect-17.4)
that if we consider integrals in the sense of usual improper integrals they will not converge as $\omega=\pm \beta$ while in the sense of the principal value they will (not required).