Numerical evaluation of integral gives warning message

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Gunnar 2018-6-27

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评论： Walter Roberson 2018-6-28

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Dear all,

I am trying to compute the following integral in MATLAB:

with corresponding code is:

p1_minusSign = @(r) exp(-2*pi * integral(@(x) x/(1-x^4/r^4), r, Inf, 'ArrayValued', 1)) * exp(-pi*r^2)*2*pi*r;
p2_minusSign = integral(p1_minusSign, 0, Inf, 'ArrayValued', 1),

Unfortunately, MATLAB gives me a bunch of warning messages of the following form: "Warning: Minimum step size reached near x = 5833.72. There may be a singularity, or the tolerances may be too tight for this problem."

As far as I can tell, I don't see any potential singularity in my expression. In contrast, if we just change the sign in the demoninator of the inside integral, it computes perfectly:

with corresponding code:

p1_plusSign = @(r) exp(-2*pi * integral(@(x) x/(1+x^4/r^4), r, Inf, 'ArrayValued', 1)) * exp(-pi*r^2)*2*pi*r;
p2_plusSign = integral(p1_plusSign, 0, Inf, 'ArrayValued', 1),

Any ideas on what the error might be and how to correct it?

Many thanks in advance for your help.

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Torsten 2018-6-27

Shouldn't the limits of the integral in the exp(...) expression be x=0 to x=r instead of x=r to x=Inf ?

Gunnar 2018-6-28

@Torsten: No, the limits are fine I think. Basically, with respect to a a point in 2D, we are integrating with respect to an outer region from r to Inf, and an inner region from 0 to r.

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回答（2 个）

John D'Errico 2018-6-27

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编辑：John D'Errico 2018-6-27

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Ignoring the fact that you cannot just factor out -1 from that term in the integral, and get what you think you got, we have the added problem that your inner integral is not even finite.

syms x r
K = x/(1 - x^4/r^4)
K =
-x/(x^4/r^4 - 1)

Now, pick some arbitrary value for r. Lets say 1.

int(subs(K,r,1),[1,inf])
ans =
-Inf

But if you change the denominator in the simple way that is NOT mathematically correct, as you wrote, then it is of course finite, as expected.

K2 = x/(1 + x^4/r^4)
K2 =
x/(x^4/r^4 + 1)
int(subs(K2,r,1),[1,inf])
ans =
pi/8

There is a fundamental difference between those two kernels. So, K2 is a mountain that we can indeed climb. (Sorry about that. Could not resist it.) It would probably be a HW assignment in first year calc, to show the former case is not finite?

Anyway, the answer is your problem lacks a finite solution.

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John D'Errico 2018-6-28

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I might accept that your double integral could possibly exist, even though the inner integral does not. Part of me worries if that can be true. But I certainly cannot argue that it will never be possible to resolve this problem without some significant thought on the matter. All I can do at the moment is argue that you cannot solve the problem as you have tried to formulate it.

The approach of constructing a piecewise solution does not apply here. For example, some integration kernels can have a singularity in them, yet still have a finite integral. A beta function is a simple case. Given the correct set of arguments, the beta function kernel can be singular at the end point, yet have a finite integral. Thus we can write this

int(x^(-1/2),[0,1])
ans =
2

and get the expected finite result, even though a numerical integration tool might balk at the singularity of the function on that domain. The integral still exists there as a finite value.

In your problem however, as it is written, the inner integral has no finite result. The problem is not that the numerical integration tool cannot resolve the issue. It is that for any value of r, the inner integral apparently does not exist.