How to solve an integral with the Inverse error function

Click For Summary

Discussion Overview

The discussion revolves around the challenge of solving a specific integral involving the inverse error function, particularly focusing on analytical approximations and numerical solutions. The integral is defined with parameters that introduce complexity, especially as the variable approaches a critical value.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents an integral involving the inverse error function and expresses difficulty in finding an analytical approximation, especially near a critical point where divergence occurs.
  • Another participant suggests using a Taylor series to approximate the integrand.
  • A participant questions the feasibility of Taylor expanding at the point of divergence and seeks ways to normalize the integral to simplify numerical solutions.
  • Another participant proposes that the inverse error function can be expressed as a Maclaurin series, though they are uncertain if this is the best approach.
  • One participant expresses a desire to express the integral in a form that separates variables, potentially simplifying the integration process.
  • A later reply inquires about any progress on computing the integral, indicating ongoing interest in the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a method to solve the integral, and multiple competing approaches are discussed without resolution.

Contextual Notes

The discussion highlights challenges related to divergence in the integrand and the complexity introduced by the parameters involved, particularly the dependence on beta. There are unresolved aspects regarding the normalization of the integral and the effectiveness of proposed series expansions.

Who May Find This Useful

Readers interested in advanced mathematical techniques for solving integrals, particularly those involving special functions like the inverse error function, may find this discussion relevant.

matteo86bo
Messages
56
Reaction score
0
Hi,
this is not a homework and my problem is much bigger for me to give full details here. I came across this integral

\mathcal{I}(\xi)=\int^{\xi_c}_{\xi}{\rm d}\xi^\prime\exp\left[\sqrt{2}\sigma\,{\rm Erf}^{-1}\left(1-\frac{8\pi}{3}{\xi^\prime}^3\right)\right]


where Erf^{-1} is the inverse error function and

\xi_c=\left[\frac{3}{8\pi}\left(1-{\rm Erf}\left(\frac{\sigma^2-\sqrt{2}\sigma\,{\rm Erf^{-1}(2\beta-1)}}{\sqrt{2}\sigma}\right)\right)\right]^{1/3}

with 0\le\beta\le1.

I would like get an analytical approximation but I can't figure out a way to do that, even with software like Mathematica. I tried solving the integral numerically and I find a reliable solution, however, I'm mostly interested in points where \xi\to\xi_c^-, and here the inverse error function diverges.

Do you have any ideas on how to approximate this integral?
 
Physics news on Phys.org
Use a taylor series?
 
Thank you for your answer. But I don't understand how I can Taylor expand where the integrand diverges.
Also, do you see a way to normalize the integral and/or make it more simple to solve it numerically.
What I find challenging is that for every value of beta I have to numerically solve the integral. It would be better to normalize somehow the result with beta and then do the integral only once.
 
The inverse errorfunction can be defined as a Maclaurin Series.
That seems like a way to get rid of some awkwardness there, but I don't know for sure that it is the best way to go.

But I don't immediately see how to make it much simpler besides that.
 
Thanks, the problem seems very complicated and I think I have to resort to a numerical integration.
However, that would be easier if one could write

\mathcal{I}(\xi,\beta)=f(\beta)\mathcal{I}^{\prime}(\xi)

so this way I would have to integrate only once. Do you see a way to possibly achieve that?
 
I understand this is quite an old post but: any luck with computing this integral? I have come across something similar recently.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K