What does this integral represent?

  • I
  • Thread starter Anixx
  • Start date
  • #1
62
4
I cannot understand what this integral is doing:

$$g(x)=\left(\frac{i \pi}{2}-\gamma\right) f(x)+\frac{1}{2}\,\text{P.V.}\int_{-\infty}^\infty \left(\frac{1}{x-x'}-\frac{1}{| x-x'| }\right)\,f(x')\,dx'$$

Can anybody please rewrite it in a more understandable form?
 

Answers and Replies

  • #2
13,020
6,907
What is the context for this integral?

is this from a physics problem?

What does the P and V represent?

is this homework? For what class? What book?
 
  • #3
haushofer
Science Advisor
Insights Author
2,543
939
I guess Principal Value; the integral diverges for x=x'. But some more context would be nice; I left my crystal ball at home today.
 
  • #5
jasonRF
Science Advisor
Gold Member
1,462
518
To me this looks like a distribution theory question involving some pretty standard pseudofunctions, but like the other responders I would want more info from Annix before crafting a reply that could easily be useless.

In addition to the info that jedishrfu requested, I would like to know the properties of ##f(x)##. Is it continuous? How does it behave as ##x \rightarrow \pm \infty##?

jason
 
  • #6
319
108
$$\text{P.V.}\int_{-\infty}^\infty \left(\frac{1}{x-x'}-\frac{1}{| x-x'| }\right)\,f(x')\,dx'$$

Maybe would be interesting to ask the question, "Come up with a non-zero f(x) that would make this integral converge in the principal-value sense?" Just looking at it at face-value, f(x) would need to result in an expression I think of the indeterminant form ##\infty-\infty## which would then converge in the P.V. sense. Off hand I can't think of a function for this.
 
  • #7
jasonRF
Science Advisor
Gold Member
1,462
518
The integrand is zero when ##x^\prime < x##, so I don't think this is a principle value integral. It has at least one singularity and possibly more, depending on ##f##. I have a reasonable guess as to what the integral might represent (assuming ##f## is nice enough), but my guess could be wrong. Hence the wait for more information.

jasn
 
  • #9
jasonRF
Science Advisor
Gold Member
1,462
518
I don't know if the formula is what you want, but it seems that they are indeed working with distributions (generalized functions). The use of the ##P.V.## notation is what made it very confusing since there is no principle value here. To me it appears the finite part of the integral is what is desired, as is done when defining pseudofunctions in distribution theory. If the notation had been ##Pf## (for pseudofunction) or ##Fp## (for finite part) it would have been clear, although to be fair some authors (eg Gelfand) don't use any special notation to indicate such things. Assuming this interpretation, I first rewrite this as the simpler
$$
\begin{eqnarray*}
\frac{1}{2} Fp \int_{-\infty}^{\infty} \left(\frac{1}{x-x^\prime} - \frac{1}{\left|x-x^\prime\right|}\right) f(x^\prime) \, dx^\prime & = & - Fp \int_x^\infty \frac{f(x^\prime)}{x^\prime-x} dx^\prime
\end{eqnarray*}
$$
where I am using the symbol ##Fp## to denote the fact that integral as written diverges but that we will be regularizing it to make it convergent. Now we use the standard approach and define the singular distribution above as some finite-order derivative of a regular distribution. In this case, since for normal functions ##\frac{d}{dx^\prime}\ln(x^\prime-x)=1/(x^\prime-x)##, we just need to compute the distributional derivative of the distribution that is ##\ln(x^\prime-x)## when ##x^{\prime}>x## and zero otherwise. So we have
$$
\begin{eqnarray*}
Fp \int_x^\infty \frac{f(x^\prime)}{x^\prime-x} dx^\prime & = &- \int_x^\infty \ln(x^\prime-x) \frac{df(x^\prime)}{dx^\prime} \, dx^\prime \\
& = & -\lim_{\epsilon\rightarrow 0^+} \int_{x+\epsilon}^\infty \ln(x^\prime-x) \frac{df(x^\prime)}{dx^\prime} \, dx^\prime \\
& = & \lim_{\epsilon\rightarrow 0^+} \left[f(x) \, \ln\epsilon + \int_{x+\epsilon}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime\right] \\
& = & \lim_{\epsilon\rightarrow 0^+} \left[-f(x)\int_{x+\epsilon}^{x+1} \frac{1}{x^\prime-x} dx^\prime + \int_{x+\epsilon}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime \right] \\
& = & \lim_{\epsilon\rightarrow 0^+} \int_{x+\epsilon}^{x+1} \frac{f(x^\prime)-f(x)}{x^\prime-x} dx^\prime + \int_{x+1}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime \\
& = & \int_{x}^{x+1} \frac{f(x^\prime)-f(x)}{x^\prime-x} dx^\prime + \int_{x+1}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime
\end{eqnarray*}
$$
Here the first line simply uses the definition of the distributional derivative (it should be familiar to anyone who has seen how the derivative of the delta function behaves), the third line results from integration by parts, and the fourth line is simply rewriting the logarithm in a useful way. Besides the first integral with the ##Fp## in front of it, every expression above is convergent and all of the steps are justified as long as ##f## is "nice" enough. For sure if ##f## is a Schwartz function (https://en.wikipedia.org/wiki/Schwartz_space) then this holds, but a weaker set of requirements should also be sufficient.

jason
 
  • #10
62
4
Thanks! Indeed it seems, Hadamard finite part was meant.

Anyway, it seems, regularization is at play here, so the exponent of the resulting operator would not be the derivative operator.

More generally, they get
$$
\ln^? (D)=\left( \begin{array}{ccccccc} \psi (1) & 0 & 0 & 0 & 0 & 0 & \cdot \\ 0 & \psi (2) & 0 & 0 & 0 & 0 & \cdot \\ 0 & 0 & \psi (3) & 0 & 0 & 0 & \cdot \\ 0 & 0 & 0 & \psi (4) & 0 & 0 & \cdot \\ 0 & 0 & 0 & 0 & \psi (5) & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & \psi (6) & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{array} \right)-\ln x
$$
or
$$
\ln^? (xD)=\left( \begin{array}{ccccccc} \psi (1) & 0 & 0 & 0 & 0 & 0 & \cdot \\ 0 & \psi (2) & 0 & 0 & 0 & 0 & \cdot \\ 0 & 0 & \psi (3) & 0 & 0 & 0 & \cdot \\ 0 & 0 & 0 & \psi (4) & 0 & 0 & \cdot \\ 0 & 0 & 0 & 0 & \psi (5) & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & \psi (6) & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{array} \right)
$$

while the true, verifiable logarithm of ##xD## is
$$\ln(xD)=\left( \begin{array}{ccccccc} -\infty & 0 & 0 & 0 & 0 & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & 0 & . \\ 0 & 0 & \ln 2 & 0 & 0 & 0 & . \\ 0 & 0 & 0 & \ln 3 & 0 & 0 & . \\ 0 & 0 & 0 & 0 & \ln 4 & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & \ln 5 & . \\ . & . & . & . & . & . & . \\ \end{array} \right)$$


See here: https://math.stackexchange.com/ques...ations-of-ln-xd-operator-which-one-is-correct

Interestingly enough, their operator can be obtained in terms of regularization of divergent integrals:

$$\text{reg }\ln(\omega_++xD)=\ln^? (xD)$$

Still, I wonder where the regularization comes from in the original article. It seems, they did not use finite part.

 

Related Threads on What does this integral represent?

Replies
1
Views
4K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
645
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
7
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
Top