# Fourier Transform: Why is this so?

1. Aug 18, 2008

### rsq_a

Why is this so?

$$\displaystyle F\left[ \frac{1}{1-e^{-\pi x}} \right] = i \frac{1+e^{-2k}}{1-e^{-2k}}$$

Here, $$-\infty < x < \infty$$. It has to be done by contour integration, by the way. Unfortunately, I'm having difficulty with the whole thing.

2. Aug 18, 2008

### jostpuur

Are you sure that is so? To me it seems that the Fourier transform isn't defined for that function. The integral diverges at x=0.

3. Aug 18, 2008

### rsq_a

Yes, I'm (99%) sure.

The function diverges (not only at x = 0, but also at x = infinity), but not faster than $$e^{\alpha x}$$. That should be enough to assure existence of a Fourier transform.

4. Aug 18, 2008

### jostpuur

If I try to keep things simple, then the situation is that the integral

$$\int\limits_{-\infty}^{\infty} \frac{e^{ikx}}{1-e^{-\pi x}} dx$$

does not converge, and the Fourier transform doesn't get defined. I don't know what rate of divergence could mean in this context.

Actually there is two divergence problems. One is at $$x=0$$, and other one at $$x=\infty$$.

It could be there is something going on that I'm not aware of. I'm probably not wrong to guess that this has something to do with some regularizations?

5. Aug 18, 2008

### rsq_a

To be perfectly honest, I'm not THAT sure of anything.

The standard idea I've known is to instead examine the fourier transform of $$e^{\alpha x}f(x)$$. When one does the fourier inversion, it amounts to integrating not on the real k axis, but along the line $$[-\infty + i\alpha,-\infty + i\alpha]$$ (there may be a sign off there). This is why we don't need functions that have convergent integrals -- so long as they don't shoot off faster than an exponential.

The problem is that this gives you the fourier transform $$F(k+\alpha)$$, not F(k). It of course isn't a problem once you invert (you get f(x) in terms of an integral -- just not the one on the real k-axis). But the actual F(k) expression...?

Indeed, when I went down this line of thought, my impression was F(k) doesn't exist. But this is a published result. Moreover, the rest of the results makes sense, which is why I don't doubt this claim. The author writes that it follows from contour integration. No other hint is given.

6. Aug 18, 2008

### morphism

What is your definition of "Fourier transform"?

7. Aug 18, 2008

### rsq_a

Standard definition. Possibly extended to the complex plane.

8. Aug 18, 2008

### rsq_a

Nevermind. It's been solved at SOS.

Idiot mistake I made not remembering $$1-e^{\pi z}$$ has more than the one root at z = 0.

9. Aug 18, 2008

### jostpuur

What does the notation P.V. mean?

10. Aug 18, 2008

### rsq_a

http://en.wikipedia.org/wiki/Cauchy_principal_value

Basically, the integral without the singularity. The poster misused some of the notation. The fourier integral is not defined into a principal value, but as "the integral over the washer". It's by virtue of the fact the outer radial integral tends to zero that the principal value is equal to the residues and thus equal to the fourier integral (modulo signs and stuff).

But of course, I'm not complaining. I got my answer.

11. Aug 18, 2008

### jostpuur

Ok. It should be stated in the original problem, that the divergence at the origo shall be dealt with the Cauchy principle value. I would be curious to know how reliable the inverse Fourier transform is, once the Fourier transform has been calculated with Cauchy principle value choice. You know, you can deal with the divergence in many different ways, and the inverse formula is not going to hold for them all, at least.

12. Aug 18, 2008

### jostpuur

I'm still skeptical over this entire result. The expression

$$\frac{e^{-ikz}}{1 - e^{-\pi z}}$$

does not approach zero on the limit $$z\to (+\infty, 0)$$, and does not oscillate rapidly either to make the integral go zero. So the vanishing of the integral on the outer semi-circle looks suspicious too.

13. Aug 18, 2008

### rsq_a

That's standard...

The Fourier transform was never calculated with just the Cauchy principal value (as I said in my post above, correcting the solution).

It was calculated by applying the definition: an integral from x = -infinity to +infinity. The contour started at -infinity, ran along the x-axis (this is the principal value), plus the indentations above the singularities, and over the outside of the "washer". It just so happens that the function decays in the outside washer.

There's no ambiguity here. It's the standard way of dealing with fourier transforms in the complex plane.

14. Aug 18, 2008

### rsq_a

As z tends to infinity, the integrand behaves like exp(-ik + pi)z. Thus for Im(k) < -pi, the integral tends to zero. I think that's enough, but I'm not quite sure about the portion near the real-axis. Gotta go, but I'll think about it.

15. Aug 18, 2008

### jostpuur

$$\lim_{z\to (+\infty, 0)}\frac{1}{1-e^{-\pi z}} \to 1$$

so the integrand behaves like

$$\frac{e^{-ikz}}{1-e^{-\pi z}} \to e^{-ikz}.$$

There is no chance this integral is converging, and also no chance that the residue trickery could have been valid.

You could interpret the integral as a distribution though, like first fixing the integration domain as finite, and then taking the limit when the domain extends to all real axis, and solving how the integral function behaves as a distribution on this limit.

16. Aug 18, 2008

### rsq_a

You're right. I was too hasty writing what I wrote when I wrote it. Here is a reply, but it may have an error. I'm in a hurry though.

Define,

$$f_{+}(x) = \frac{1}{1-e^{-\pi x}}$$ for $$x > 0$$ and,

$$f_+ = 0$$ otherwise.

Let $$F_+(x) = e^{\alpha x}f_+(x)$$ where $$\alpha < 0$$ and real.

Now calculate,

$$\hat{F_+}(k) = \textbf{F}[F_+]$$

Notice now the integrand tends to zero for,

$$\text{Im}(k) = 0$$ and $$z \to \infty$$

By the same method as above,

$$\hat{F_+}(k) = i\left(\frac{1+e^{-2(k+\alpha i)}}{1-e^{-2(k+\alpha i)}}\right)$$

It is consistent (by extension of the Inverse Fourier Transform to the complex domain) to define,

$$\hat{F_+}(k) = \hat{f_+}(k+i\alpha)$$

Thus,

$$\hat{f_+} =i\left(\frac{1+e^{-2x}}{1-e^{-2x}}\right)$$

The same follows for $$f_-(x)$$ defined similarly and the result follows from superposition.

17. Aug 19, 2008

Where I come from, what you're talking about is called the Laplace transform. The idea is that you replace the $i\omega$ in the Fourier transform with a complex variable $s=\sigma+i\omega$. Then, the transform integral is exactly the Fourier transform of $e^{\sigma x}f(x)$. In this setting, $\sigma$ is also considered a variable, and so you get a function of $s$, along with an associated region of convergence (typically a range of $\sigma$ values that the integral converges for). The $\sigma$ component is sort of "extra" information in the sense that the time-domain function can be recovered by a line integral along $\omega$ for any value of $\sigma$ in the region of convergence, as you are discussing here.

It bears emphasizing that when the $i\omega$ axis is not included in the region of convergence, the Fourier transform does not, strictly speaking, exist. The Laplace transform will exist, though, provided f(x) can be made convergent through exponential damping, as you say. Note that, for example, an f(x) that blows up for both large positive and large negative x cannot be made to converge, as the end will still blow up with $\sigma$ positive, and vice-versa with $\sigma$ negative. You can't get around this by chopping it into two pieces in the time domain, as the corresponding regions of convergence of the transforms will not overlap, and so there is no way to put the two transforms back together.

18. Aug 19, 2008

### rsq_a

Well, not exactly the Laplace Transform. The Laplace Transform is the Fourier Transform for the $f_+$ I defined, with $k = ip$ and thus the inversion contour inversion is taken along $\alpha -i\infty$ to $\alpha +i\infty$ -- a vertical line instead of a horizontal one. Thanks for pointing this connection out

Hmm...Interesting. I never thought about that.

But what if I did tried this for the fourier transform of $f(x) = e^{|x|}$, clearly divergent at both ends.

Examine $F(x) = e^{-|2x|}f(x)$ and define as before $F_{\pm}(x)$ as the chopped-in-two-pieces version with the other side set to zero.

Then for $\hat{F}(k_1 + ik_2)[/tex] is defined (I believe(?)) for [itex]x>0$ when $k_2 > -1$ and for $x<0$ for $$k_2 < 3$$ -- an overlapping strip.

Why would this not work?

19. Aug 19, 2008

I'm not sure I see the difference between that and a Laplace transform...?

Well in this case, you wouldn't even need to chop the function apart, right? You'd be taking the Fourier transform of $e^{-|x|}$, which is a convergent function. The trouble with this approach is that the transform of $e^{-|\sigma x|}f(x)$ is not related to the transform of $f(x)$ in a nice way, as is the case for the transform of $e^{-\sigma x}f(x)$. So it's not clear to me that this approach would get you very far, outside of very specific functions like $f(x)=e^{-|ax|}$...

20. Aug 19, 2008

### rsq_a

Yes, but your original statement was that we can't take fourier transforms of functions which have divergence at both tails.

I was wondering whether the statement was a pragmatic one or a general one.