# Fourier Transform Convergence

1. Dec 8, 2013

### nabeel17

From -infinity to infinity at the extreme ends do Fourier transforms always converge to 0? I know in the case of signals, you can never have an infinite signal so it does go to 0, but speaking in general if you are taking the fourier transform of f(x)

If you do integration by parts, you get a term (f(x)e^ikx evaluated from -infinity to infinity why does this always = 0?

2. Dec 8, 2013

### Staff: Mentor

No, not always. If a signal is periodic in one domain then it is discrete in the other domain. So if you have a signal which is discrete in time, then it is periodic in frequency. Since it is periodic in frequency it does not converge to 0 at infinity.

3. Dec 8, 2013

### nabeel17

Ok then why is it that we the first term in the integration by parts goes to 0 then regardless of the function (Whether it is periodic or not)? For example when finding the fourier transform of a derivative F[d/dx] = ∫d/dxf(x)e^ikx= f(x)e^ikx evaluated -infinity to infinity -ik∫f(x)e^ikx

the first term = 0, why is that? If it were a wave function like in QM then it makes sense because the area under the wave function must be finite and converge to 0 at the extremes for it to have a probability density, but why here?

4. Dec 8, 2013

### Staff: Mentor

I think that the various properties of the Fourier transform all assume that f satisfies the Dirichlet conditions.

5. Dec 8, 2013

### AlephZero

The OP is asking about Fourier transforms, not Fourier series (of periodic functions) which is what #2 and #4 appear to be about.

A reasonable condition for Fourier transforms to behave sensibly is that $\int_{-\infty}^{+\infty}|f(x)|dx$ is finite. Note that if you use Lebesque measure to define integration, that does not imply $f(x)$ converges to 0 as x tends to infinity. $f(x)$ can take any values on a set of measure zero.

(Also note, "reasonable" does not necessarily mean either "necessary" or "sufficient"!)

The mathematical correspondence between Fourier series and Fourier transforms is not quite "obvious", since the Fourier transform of a periodic function (defined by an integral with an infinte range) involves Dirac delta functions, and indeed the Fourier transform of a periodic function is identically zero except on a set of measure zero (i.e. the points usually called the "Fourier coefficients").

On the other hand if you integrate over one period of a periodic function, it is a lot simpler to get to some practical results, even if you have to skate over why the math "really" works out that way.

Last edited: Dec 8, 2013