What is the value of the integral for higher order poles in the real axis?

[Santilopez10]

Hello! I have been searching the web and textbooks for a certain theorem which generalizes the value of the integral around a infinitesimal contour in the real axis, or also called indented contour over a nth order pole.
It is easy to prove that if the pole is of simple order, the value of the integral is just $-i\pi Residue(f(z_0))$, but problem arises when $n\geq2$. for example: $$ \oint_C \frac {e^{iz(x-1)}}{z^2}\, dz $$
Which arises in the search for the inverse Fourier transform of the Fourier transform of x, in the range 0<x<1, which basically is proving the duality of the Fourier transform for the defined function.
Now, if our contour has a indented path at z=0, we make the substitution $z=\epsilon e^{i \phi}, dz=i \epsilon e^{i \phi} d\phi$ and we let $\phi$ run from $\pi$ to 0, we get: $$ \int_\pi^0 \frac {e^{i\epsilon e^{i\phi} (x-1)}}{\epsilon^2 e^{2i\phi}} i\epsilon e^{i\phi}\, d\phi$$
which isn`t defined if we let $\epsilon \rightarrow 0$. So I am stuck there, the integral should exist, as it does have a defined inverse Fourier transform in the range 0<x<1. So the question is: Does anyone know any theorem which states the value of the integral for higher order poles in the real axis?

EDIT: The function is square integrable, so the inverse Fourier transform must exist.
the origal Fourier transform is $\frac {e^{-iw}(iw-e^{iw}+1)}{w^2}$

 

[jasonRF]

The Fourier transform is analytic at $\omega=0$. Your inversion integral does not require an indented path.

 

[Santilopez10]

The Fourier transform is analytic at $\omega=0$. Your inversion integral does not require an indented path.

mind showing why it is analytic? I´ve found that it has a simple pole at 0, which means it isn´t analytic at w=0.

 

[jasonRF]

Try Taylor expanding the numerator about $\omega=0$; it should fall out pretty easily that there is no pole at $\omega=0$.

By the way, in general the Fourier transform of a time-limited, square-integrable function is analytic in the entire complex plane.

Jason

EDIT: Also, you should be able to take the limit of your Fourier transform as $\omega\rightarrow 0$ and show that it is finite, which shows that there is no pole at the origin.

 

[Santilopez10]

Try Taylor expanding the numerator about $\omega=0$; it should fall out pretty easily that there is no pole at $\omega=0$.

By the way, in general the Fourier transform of a time-limited, square-integrable function is analytic in the entire complex plane.

Jason

EDIT: Also, you should be able to take the limit of your Fourier transform as $\omega\rightarrow 0$ and show that it is finite, which shows that there is no pole at the origin.

Yes, you are right, is has value 0 at z=0, so it is analytic there, still, which contour would you recommend to find it´s inverse Fourier transform?

 

[jasonRF]

Actually, I get that the Fourier transform is 1/2 at the origin.

Anyway, if $f(t)$ is the original function and $F(\omega)$ is its Fourier transform, the inversion theorem tells us that
$$ \begin{eqnarray}
f(t) & = & \lim_{R\rightarrow \infty} \frac{1}{2\pi} \int_{-R}^{R} F(\omega) \, e^{i\omega t} \, d\omega\\
& = & \lim_{R\rightarrow \infty} \frac{1}{2\pi} \int_{-R}^{R} \frac{e^{i\omega (t-1)} \left(i\omega+1 \right) - e^{i\omega t}}{\omega^2} \, d\omega
\end{eqnarray}
$$

Since the the integrand is entire we are free to deform the contour to be anything we want. There are two contours that we can try: 1) use a semicircle of radius $R$ in the lower half-plane. As $R\rightarrow \infty$ you should show that this contour (and hence $f(t)$) will vanish for some values of $t$. 2) use a semicircle of radius $R$ in the upper half plane. Again, as $R\rightarrow \infty$ you should show that this contour (and hence $f(t)$) will vanish for some other values of $t$.

However, there will still be a set of values for $t$ remaining for which we cannot easily show $f(t)=0$ (since you know the original function you know it is $0\leq t \leq 1$). I haven't really thought about how to do that integral - it doesn't look easy on the surface.

Jason

 

[jasonRF]

By the way, I was a little sloppy. The inversion theorem actually says,
$$\begin{eqnarray}
\frac{f(t^+) + f(t^-)}{2} & = & \lim_{R\rightarrow \infty} \frac{1}{2\pi} \int_{-R}^{R} F(\omega) \, e^{i\omega t} \, d\omega.
\end{eqnarray}$$
The left side of the equation simply says that if $f$ is discontinuous, the inverse transform will give you the average value of $f$ as you approach the discontinuity from each side. For your function, the inversion integral should give you the value 1/2 when $t=1$.

 

[Santilopez10]

yeah, you were right, it´s actually 1/2 at z=0, just checked, I´ve tried separating the original integral in 2 different integrals and trying different contours but still I get nothing, shame there isn´t something similar in the internet, at least from what I have searched.

And referring to the second post, it seems reasonable that it has the value 1/2 at t=1, as it is similar to the discontuinity of a heaviside step function at 0, which has average value 1/2 as well.

 

[jasonRF]

What options you have depend on what mathematical tools you are comfortable using. If you really want to do the inverse transform as an integral, then you need to be clever. Also, I don' think you can split up the integrand into the sum of the three terms, because each of those terms isn't integrable. If you manage to do this integral, please post your solution. It will be interesting to see how you do it.

If it were me, I wouldn't try to evaluate the inversion integral at all. The tools I would use are typically learned in a course or book on Signals and Systems. Basically, I would use properties of Fourier transforms and the Fourier transforms of known functions. The tricky part is that I would actually split the integrand into the sum of the three terms. Why can I do this, when above I said that each term by itself isn't integrable? Because I am doing what all electric engineers do and allow functions and their Fourier transforms to be generalized functions; note that electric engineers usually do not explicitly state that this is what they are doing, so it can be confusing sometimes.

A good example is the Fourier transform that you need for this problem. Consider the function $\mathrm{sign}(t)$, where $\mathrm{sign}(t)=1$ for $t>0$ and $\mathrm{sign}(t)=-1$ for $t<0$. If you try to use the standard integral definition for the Fourier transform you will find that the integral does not converge. However, (and this can be made rigorous), you can allow the Fourier transform to be a generalized function and you find that the Fourier transform of $\mathrm{sign}(t)$ is $\frac{2}{i\omega}$. If you do this rigorously then you need to add some details about how to handle the singularity at $\omega=0$, but engineers usually don't worry about it.

There are then two properties that you would need. First some notation: let $f(t)$ be an arbitrary function and $\mathcal{F}\left[ f(t) \right] = F(\omega) = \int_{-\infty}^{\infty} e^{-i\omega t} f(t)\, dt$ its Fourier transform. Then the properties you need are
$$\begin{eqnarray*}
\mathcal{F}\left[ f(t-\tau) \right] & = & e^{-i \omega \tau} F(\omega) \\
\mathcal{F}\left[t f(t) \right] & = & i \frac{d}{d\omega} F(\omega)
\end{eqnarray*}$$

Here is an example for one term, where I find the inverse transform of $e^{-i\omega} /\omega^2$. Start with
$$ \mathcal{F}\left[ \mathrm{sign}(t) \right] = \frac{2}{i \omega}.$$
Now we try to use the properties above to make the right hand side look like the our desired transform. First, use the second property above to get,
$$ \mathcal{F}\left[t\, \mathrm{sign}(t) \right] = i \frac{d}{d\omega} \frac{2}{i \omega} = - \frac{2}{\omega^2}.$$
We are almost there. Now use the first property above,
$$ \mathcal{F}\left[(t-1) \,\mathrm{sign}(t-1) \right] = - e^{-i\omega} \frac{2}{\omega^2}.$$
Finally, just divide both sides by $-2$,
$$ \mathcal{F}\left[\frac{1}{2}(1 - t)\, \mathrm{sign}(t-1) \right] = e^{-i\omega} \frac{1}{\omega^2}.$$
Letting $\mathcal{F}^{-1}$ denote the inverse transform, we have
$$\mathcal{F}^{-1}\left[ e^{-i\omega} \frac{1}{\omega^2} \right] = \frac{1}{2}(1-t)\, \mathrm{sign}(t-1) .$$

The other terms can then be found using the same approach. If you work through this, it would be great if you can show us your work.

Jason

 

[Santilopez10]

actually I was thinking about splitting into 2 integrals, which would be like this: $$\frac {1}{2\pi} (i \oint_C \frac {e^{iz(x-1)}}{z} \, dz + \oint_C \frac {e^{iz(x-1)}-e^{izx}}{z^2} \, dz) =\frac{1}{2\pi}(I_1+I_2)= \mathfrak {F^{-1}}(f(z))$$
The first integral is rather common and solvable with a semicircle contour using an indented contour, whereas $I_2$ is hard considering that for $0<x<1$ we need to close the contour for the separate integrals, to show that they do not diverge at the semicircle path. Considering it has a simple pole this time at z=0, we would need to compute it`s cauchy principal value, or use a contour which encloses the contour, but well, that is the hard part.