A question about a complex integral

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I was trying to calculate an integral of form:
$$\int_{-\infty}^\infty dx \frac{e^{iax}}{x^2}$$

using contour integration, with ##a>0## above. So I would calculate a contour integral with contour being a semicircle that goes along the real axis, closing it in positive direction in the upper half-plane, which would ensure exponential decay on that arc. Hence, the contour integral would be equal to the integral along the real axis.

Now to remedy the singularity at zero, I can change the integrand to a function: ##f(z) = \frac{e^{ix}}{(x\pm i\epsilon)^2}##. Here the plus sign would shift the pole from the real axis to the lower half-plane, and the minus sign would shift it to the upper half-plane. Then I would take the limit as ##\epsilon \rightarrow 0##. However, if I shift the pole to the lower part of the plane, it would go outside the contour, hence the contour integral would be equal to zero by Cauchy residue theorem. But if I shift it into the upper part of the plane, it would go inside the contour, and the residue is calculated to be ##iae^{-a\epsilon}##, which isn't zero.

I don't understand what I'm doing wrong here, as it seems that my choice of regularization changes the value of the integral, which shouldn't happen.
 

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  • #2
pasmith
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You need to replace the integral over the real axis with integrals over [itex][-R,-\epsilon][/itex] and [itex][\epsilon, R][/itex] and an integral over a semi-circle of radius [itex]\epsilon[/itex] in the opposite half-plane to the semi-circle of radius [itex]R[/itex]. And then you need to show that the integral over the small semi-circle vanishes as [itex]\epsilon \to 0[/itex], which may not be the case.
 
  • #3
Svein
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Expand [itex]e^{iax} [/itex] as a power series: [itex]1+\frac{iax}{1!}+\frac{(iax)^{2}}{2!}+\dotso [/itex]. Divide through by [itex]x^{2}[/itex]: [itex]\frac{1}{x^{2}}+\frac{ia}{1!\cdot x}+\frac{(ia)^{2}}{2!}+\dotso [/itex]. This gets you the residue at x=0 (the coefficient at [itex]x^{-1} [/itex]) as [itex]ia [/itex]. This means the the integral [itex] \int_{C}\frac{e^{iaz}dz}{z^{2}}[/itex] where C is any circle that encloses zero is [itex]2\pi i\cdot ia = -2\pi a [/itex]. Now show that the integral over the lower half of C tends to the half of that value as the radius of C tends to 0.
 
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  • #4
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Okay, @Svein, @pasmith, I will attempt to calculate the integral over the semicircle before taking the limit of the radius to zero. So, if I take this semicircle around zero, I can parametrize it with ##z = \epsilon e^{i\theta}##, where ##\theta \in [-\pi,0]##.
Then:
$$\int_C \frac{e^{iaz}dz}{z^2} = i\epsilon\int_{-\pi}^0 d\theta e^{i\theta}\frac{e^{ia\epsilon e^{i\theta}}}{\epsilon^2 e^{2i\theta}} = \frac{i}{\epsilon}\int_{-\pi}^0 d\theta e^{-i\theta}\sum_{m=0}^{\infty}\frac{(ia\epsilon e^{i\theta})^m}{m!} = i\sum_{m=0}^{\infty} \frac{(ia)^m}{m!}\epsilon^{m-1}\int_{-\pi}^0 d\theta e^{i(m-1)\theta}$$
Evaluating the integral, the sum becomes:
$$\int_C \frac{e^{iaz}dz}{z^2}=-\frac{2}{\epsilon} - a\pi +\sum_{m=2}^\infty \frac{(ia)^m}{m!}\frac{\epsilon^{m-1}}{m-1}(1-(-1)^{m-1})$$
But now in the limit as ##\epsilon## approaches 0, I will indeed get half your value in the residual term, and in the sum will reduce to zero term by term, however I still have this divergence in the first term. So...that's the issue, if I didn't make a mistake anywhere above.
 
  • #5
jasonRF
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I don't understand what I'm doing wrong here, as it seems that my choice of regularization changes the value of the integral, which shouldn't happen.
What is the context of this integral? This type of behavior happens all the time when doing things like solving PDEs using integral transform techniques. As an example, for the simple wave equation with no losses you get poles on the real axis. If you shift them one way you get causal solutions, while if you shift them the other way you will find that your waves travel towards the source, not away. In that case one solution to the ambiguity is to add a small amount of loss to the problem and see where the singularities end up; another is to try both options for shifting the singularities and only keep the result that makes physical sense. The point I am making is that the math doesn't `"know" what kind of solution you want - if you haven't included everything in the equations themselves then it is up to you to work with the math to find the type of solution you are looking for.

But now in the limit as ##\epsilon## approaches 0, I will indeed get half your value in the residual term, and in the sum will reduce to zero term by term, however I still have this divergence in the first term. So...that's the issue, if I didn't make a mistake anywhere above.
Yes, that trick of taking the limit of a small circular contour works fine at a simple pole, but never at a higher order pole.

The real issue is that the integral diverges. I suspect there is nothing you can do to change that. What is the source of the integral?

Bu the way, if it is from an application where you might expect a distribution (generalized function) as a solution then there will be a pretty nice answer, but trying to evaluate the integral this way will not lead you to it.

jason
 
  • #6
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@jasonRF
The context of this integral is a detector response function in curved spacetime, basically a problem that's equivalent(physically similar) to Hawking radiation or Unruh effect or cosmological particle creation, in my concrete problem it's a free-falling detector in De-Sitter space. Anyways, this integral is the simplified version of the integral I started with, and as you noted, I can't really change anything about the divergence.

The thing is, the author of the book I'm learning this from gave the solution after calculating this integral 'using contour integration', without further explanation, so I was deriving the result. The complete expression I was looking for has the form of:
$$\sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{iax}}{(x+2i\pi k)^2}$$
So basically, we have poles on the imaginary axis, and if we're doing it with semicircle contour, we have to close it in the upper plane, since ##a>0##. And all the poles in the upper plane are caught so when those residues are calculated, you get a geometric series which sums up to a Planck type of distribution. The zeroth term is the issue, which is what I asked about above. It seems to me like the author put it out of the contour using ##i\epsilon## type of regularization, and ignored it, but while calculating, I wasn't paying attention to the sign of ##\epsilon##, as it shouldn't be of consequence here(in my opinion at least), and so I got a solution that was slightly different. That's how I noticed the problem I asked about in OP.

If there is an elegant answer in terms of distributions, I would be happy to hear it, too. I figured maybe using something like Plemelj-Sokhotski, but it feels like that only applies to simple poles, not poles of 2nd order, and I'm not fluent enough with distributions to solve it myself using them, but I would probably understand if you gave me some hints.
 
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  • #7
jasonRF
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You are asking a Fourier transform question, where the transforms are defined by
$$ \mathcal{F}\left[ f(x) \right]= F(a) = \int_{-\infty}^{\infty} dx \, e^{i a x} f(x).$$
So the inverse transform is
$$\mathcal{F}^{-1}\left[ F(a) \right] = f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} da \, e^{-i a x} F(a).$$
For your exact problem, ##f(x) = 1/x^2##, and the Fourier transform does not exist if you demand a function as an answer. However, there is a distributional Fourier transform, as long as you interpret ##1/x^2## correctly. And it is actually related to Plemelj, at least the version of Plemelj that I usually write as
$$\lim_{\epsilon\rightarrow 0^+} \frac{1}{x - i \epsilon} = \mathsf{Pv}\frac{1}{x} + i\pi \delta(x)$$
which of course means, for a 'nice' function ##\phi(x)##
$$\begin{eqnarray*}
\lim_{\epsilon\rightarrow 0^+}\int_{-\infty}^{\infty}dx \frac{\phi(x)}{x - i \epsilon} & = & i \pi \phi(0) + \lim_{\epsilon\rightarrow 0^+}\left[ \int_{-\infty}^{-\epsilon}dx \frac{\phi(x)}{x} + \int_{\epsilon}^{\infty}dx \frac{\phi(x)}{x}\right] \\
& \equiv & i \pi \phi(0) + \mathsf{Pv} \int_{-\infty}^{\infty} dx \frac{\phi(x)}{x}
\end{eqnarray*}
$$

Anyway, for many Fourier transform problems it is much easier to start with a transform you know and use properties of Fourier transforms to get the transform you want. In this case, I will start with the fact that the Fourier transform of what we usually write as ##1/x## is pretty well known. And if we know its transform, we can find the transform of its derivative. Instead of quoting the transform of ##1/x## I will 'derive' it by taking the inverse transform of a function that is proportional to the answer. This will also show us how to properly interpret ##1/x## (spoiler - it is a principal value, just like in Plemelj).

The great thing about generalized functions is that exchanging the order of limiting processes is almost always allowed. For example, we can change the order of a limit and a Fourier transform. So consider the function ##G(a) = sign(a)##, where the notation means what you think it means (##sign(a)=1## for ##a>0## and ##sign(a)=-1## for ##a<0##). But we cannot actually do the integral if we insert this function into our inverse Fourier transform definition. We therefore define ##G_\epsilon(a) = sign(a)\, e^{-\epsilon |a|}##, and use the fact that for generalized functions ##\mathcal{F}^{-1}\left[ \lim_{\epsilon\rightarrow 0^+} G_\epsilon(a) \right] = \lim_{\epsilon\rightarrow 0^+} \mathcal{F}^{-1}\left[ G_\epsilon(a) \right]##, as long as we interpret the limit as a generalized function (like we did in the Plemelj limit above). If you do the integral you should get
$$ \begin{eqnarray*}
f(x) & = & \mathcal{F}^{-1} \left[ G(a) \right] \\
& = & \lim_{\epsilon\rightarrow 0^+} \frac{1}{2\pi} \int_{-\infty}^{\infty} da \, e^{-i a x} G_\epsilon(a) \\
& = & \lim_{\epsilon\rightarrow 0^+}\, \frac{-i}{2\pi} \, \left[\frac{1}{x + i \epsilon} + \frac{1}{x - i \epsilon} \right] \\
& = & \frac{-i}{2\pi} \, \left[\mathsf{Pv}\frac{1}{x} - i\pi \delta(x) + \mathsf{Pv}\frac{1}{x} + i\pi \delta(x) \right] \\
& = & -\frac{i}{\pi} \mathsf{Pv}\frac{1}{x}
\end{eqnarray*}
$$
where we have used the Plemelj formula.

This tells us that ##\mathcal{F}[\mathsf{Pv} \frac{1}{x}] = i\pi \, sign(a)##. If we define ##\mathsf{Pf}\frac{1}{x^2} = -\frac{d}{dx} \mathsf{Pv}\frac{1}{x}## and use the fact that ##\mathcal{F}\left[ \frac{d f(x)}{dx} \right]= -i a F(a)##, then we find ##\mathcal{F}[\mathsf{Pf} \frac{1}{x^2}] = - \pi a \, sign(a) = -\pi |a|##. I think this solves your problem, but you should go through the algebra for sign errors, factors of ##\pi##, etc.

One last thing - we haven't specified how we interpret ##\mathsf{Pf}\frac{1}{x^2} ##. The key is to use the fact that for a generalized function ##h(x)##, the derivative is define by ##\int dx \, h^\prime(x) \phi(x) \equiv -\int dx \, h(x) \phi^\prime(x)##. So we have
$$\begin{eqnarray*}
\mathsf{Pf} \int_{-\infty}^\infty dx \frac{\phi(x)}{x^2} & = & \mathsf{Pv} \int_{-\infty}^\infty dx \frac{\phi^\prime(x)}{x^2}
\end{eqnarray*}$$
If you go through the math, you should get something like
$$ \mathsf{Pf} \int_{-\infty}^\infty dx \frac{\phi(x)}{x^2} = \lim_{\epsilon\rightarrow 0^+}\left[ \int_{-\infty}^{-\epsilon}dx \frac{\phi(x)-\phi(0)}{x^2} + \int_{\epsilon}^{\infty}dx \frac{\phi(x)-\phi(0)}{x^2}\right]
$$
By the way, ##\mathsf{Pf}## stands for 'pseudofunction'.

So if you are happy interpreting ##\frac{1}{x^2}## as ##\mathsf{Pf}\frac{1}{x^2}## in your original problem then this solves it. By the way, in my field (electrical engineering) we always just write ##1/x## and ##1/x^2## and completely ignore the principal value and pseudofunction labels.

jason
 
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  • #8
jasonRF
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By the way, in some sense you can also think of this as using the derivative of the Plemelj formula to get
$$
\lim_{\epsilon\rightarrow 0^+} \frac{1}{\left(x \mp i \epsilon\right)^2} = \mathsf{Pf}\frac{1}{x^2} \mp i\pi \delta^\prime(x)
$$
In fact, Bremermann (in his book Distributions, Complex Variables and Fourier Transforms) defines the pseudofunction ##\mathsf{Pf}\frac{1}{x^n}## by
$$
\mathsf{Pf}\frac{1}{x^n} \equiv \frac{1}{2} \lim_{\epsilon \rightarrow 0^+} \left[ \frac{1}{(x + i \epsilon)^n} + \frac{1}{(x - i \epsilon)^n} \right]
$$
If you want, you can blindly interpret ##1/x^2## as a pseudofunction and plug this into your integral. I think you will then get the average of the two cases you originally considered, which is probably the right answer?

My 'derivation' shows a little more what is going on under the hood, but the final should be equivalent, I think.

jason
 
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  • #9
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@jasonRF This was very instructive, thank you very much!

In my concrete problem, it turned out that the sign of ##i\epsilon## was tied to certain boundary conditions of a PDE that was solved somewhere in the earlier parts of the course, so my assumption that it was arbitrary was wrong for physical reasons. It turned out that if I followed the convention that was made, my results in the end turned fine, and the pole that I was puzzled about turned out to be naturally out of the integration contour.

However, the discussion you presented above was still very interesting to me, it helped me better understand some other problems I could connect it to, so thank you again!

Antarres
 
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