matlab not displaying an approximated value of an improper integral?

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Lahcen 2023-8-23

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https://ww2.mathworks.cn/matlabcentral/answers/2011812-matlab-not-displaying-an-approximated-value-of-an-improper-integral

移动：John D'Errico 2023-8-28

采纳的回答： Nathan Hardenberg

I want to calculate an approximate value of an improper integral:

>> fun= 4./(x.*(1-x).*(2-x).*(pi.^2+(log(x./(1-x)).^2))

>> vpa(int(fun,0,1))

then matlab displays the result:

ans =

vpaintegral(4/(x*(log(-x/(x - 1))^2 + 2778046668940015/281474976710656)*(x - 1)*(x - 2)), x, 0, 1).

how can I get the approximate value of this improper integral? thanks in advance.

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Dyuman Joshi 2023-8-23

在 MATLAB Online 中打开

I had an inkling because of the value but wasn't sure due to the format of the value -

sqrt(2778046668940015/281474976710656)
ans = 3.1416

Lahcen 2023-8-24

移动：John D'Errico 2023-8-28

Thank you al, Mr Nathan Hardenberg, Mr Dyuman Joshi and Mr Steven Lord, for your helpful responses in addressing my improper integral calculation issue on the Matlab forum. Your assistance was greatly appreciated!. Indeed, I worked by the idea of Mr Nathan Hardenberg.

I'd also like to note that the integral in question is convergent, whereas the numerical methods provided by Matlab seem to be getting stuck without yielding the desired result. It appears that there is a need for the development of more reliable numerical techniques.

Many thanks.

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采纳的回答

Nathan Hardenberg 2023-8-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2011812-matlab-not-displaying-an-approximated-value-of-an-improper-integral#answer_1292247

编辑：Nathan Hardenberg 2023-8-23

在 MATLAB Online 中打开

Here are some ideas, but please still regard my comment.

You have two variables x and p. You can not approximate numerically then.
"To approximate integrals directly, use vpaintegral instead of vpa. The vpaintegral function is faster and provides control over integration tolerances." [source]
Specify that you want to use x as your variable (int(fun, x, [a b]))
You can not numerically approximate if your denomenator gets 0 at some point (see example below)

-- EDIT --

Your problem is nr. 4. Your denomenator gets 0 to fix this you can simply choose values close to 0 and 1. See example below:

syms x

p = pi; % edited to be p = pi

fun = 4./(x.*(1-x).*(2-x).*(p.^2+(log(x./(1-x)).^2)))

fun =

vpa(int(fun, 0.001, 0.999)) % choosing values slightly above 0 and below 1

ans =

2.0701395395280001703925025323768

But maybe this is not good enough, since the solution can get quite a bit different when getting closer to the limits:

vpa(int(fun, 1e-110, 1 - 1e-13))  % choosing tighter values
ans = 2.7443512001078913067737891865659

-- EDIT END --

% Example of nr. 4 from above
syms x
f = 1/x;
Fvpaint = vpaintegral(f,x,[0 1])
Error using sym/vpaintegral
Failed precision goal. Try using 'MaxFunctionCalls'.

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Steven Lord 2023-8-23

在 MATLAB Online 中打开

Let's look at how your function behaves over the limits of integration. I had to add one extra closing parenthesis, which I did at the end. I'm not sure if that's the right place for it.

syms x

fun= 4./(x.*(1-x).*(2-x).*(pi.^2+(log(x./(1-x)).^2)))

fun =

fplot(fun, [0 1])

That certainly looks like a singularity at x = 1.

try
    vpa(subs(fun, x, 1))
catch ME
    fprintf('subs threw an error:\n%s', ME.message)
end
subs threw an error:
Division by zero.

The value at x = 0 also looks like it's potentially very, very large (if not infinite.)

try
    vpa(subs(fun, x, 0))
catch ME
    fprintf('subs threw an error:\n%s', ME.message)
end
subs threw an error:
Division by zero.

What would I get if I tried to numerically integrate the function?

f = matlabFunction(fun)
f = function_handle with value:
    @(x)4.0./(x.*(log(-x./(x-1.0)).^2+9.869604401089358).*(x-1.0).*(x-2.0))
integral(f, 0, 1)
Warning: Minimum step size reached near x = 1. There may be a singularity, or the tolerances may be too tight for this problem.
ans = 2.7246