Choosing the Contour for the Cauchy Integral in Fourier Transform of Norms

[emra1976]

Homework Statement

Compute the Fourier transform of a function of norm f(\norm{x}).

Homework Equations

\mathbb{F}{\frac{1}{1+\norm{x}}

The Attempt at a Solution

Attempt at using Cauchy theorem and the contour integral with the contour [(-R,R),(R,R+ip),(R+ip,-R+ip),(-R+ip,-R)] does not seem to work.

 

[Ray Vickson]

Homework Statement

Compute the Fourier transform of a function of norm f(\norm{x}).

Homework Equations

\mathbb{F}{\frac{1}{1+\norm{x}}

The Attempt at a Solution

Attempt at using Cauchy theorem and the contour integral with the contour [(-R,R),(R,R+ip),(R+ip,-R+ip),(-R+ip,-R)] does not seem to work.

Is x a scalar or a vector? If it is a scalar, do you mean |x| when you write norm(x)?

 

[emra1976]

Yes, you are correct. While I need to generalise it to x\in\mathbb{R}^n I would like to start with the case in which

x\in \mathbb{R}

and

\norm{x}=\abs{x}

 

[LCKurtz]

Emra1976, put ## or $$ tags around your tex (double characters on each end). You can check what it will look like by clicking the Go Advanced button and previewing your post.

 

[Mark44]

Yes, you are correct. While I need to generalise it to x\in\mathbb{R}^n I would like to start with the case in which

x\in \mathbb{R}

and

\norm{x}=\abs{x}

To use LaTeX here at PF you need to put the right tags around your expressions, as in the following.

Inline LaTeX

##x\in \mathbb{R}##
or
[itex]x\in \mathbb{R}[/itex]

Standalone LaTeX (for lack of a better name for it)

$$x\in \mathbb{R}$$
or
[tex]x\in \mathbb{R}[/tex]

 

[Mandelbroth]

Yes, you are correct. While I need to generalise it to x\in\mathbb{R}^n I would like to start with the case in which

x\in \mathbb{R}

and

\norm{x}=\abs{x}

Alright. Let's just crank it out. Let $f(x)=|x|$. Then, $\displaystyle \mathcal{F}\left[f(x)\right](\xi)=\int\limits_{(-\infty,+\infty)}|x|e^{-2\pi i x\xi}~dx$. You tried the residue theorem, you said?

 

[emra1976]

1) it should be $$f(x)=\frac{1}{1+\abs{x}}$$

2) I tried Cauchy theorem using the fact that $$f(z)=\frac{1}{1+\abs{z}}$$ is analytical. I tried both the rectangular contour $$([-R,R],[R,R+ip],[R+ip,-R+ip],[-R+ip,-R])$$ and the the countour of the semidisk centered on the origin with radius $$R$$.

 

[Ray Vickson]

1) it should be $$f(x)=\frac{1}{1+\abs{x}}$$

2) I tried Cauchy theorem using the fact that $$f(z)=\frac{1}{1+\abs{z}}$$ is analytical. I tried both the rectangular contour $$([-R,R],[R,R+ip],[R+ip,-R+ip],[-R+ip,-R])$$ and the the countour of the semidisk centered on the origin with radius $$R$$.

In TeX, just write |z|, not \abs(z).

The function z| is not an analytical function, so I don't think you can apply Cauchy directly.

 

[emra1976]

Do you mean that $$\frac{1}{1+|z|}$$ is not analytical (used as homomorphic)?
If so, where are its poles?

 

[Ray Vickson]

Do you mean that $$\frac{1}{1+|z|}$$ is not analytical (used as homomorphic)?
If so, where are its poles?

No poles, but it fails to satisfy the Cauchy-Riemann equations, so is not holomorphic.

However, you can write
$$F(w) = \int_{R} \frac{e^{-iwx}}{1+|x|} \, dx = 2\int_0^{\infty} \frac{\cos(w x)}{1+x} \, dx,$$
and then extend the latter integrand to
$$\frac{\cos(w z)}{1+z},$$
which IS analytic.

 

[emra1976]

Now, given
$$f(z)=\frac{cos(\omega z)}{1+\omega z}$$
how would you choose the contour for the Cauchy Integral? My two choices would be

1) the quarter disk in the top right quadrant of the complex plane
2) a rectangular contour in the top right quadrant of the complex plane

but I get stacked with both (1) and (2).