# matlab not displaying an approximated value of an improper integral?

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Lahcen on 23 Aug 2023
Moved: John D'Errico on 28 Aug 2023
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 on 23 Aug 2023
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 on 24 Aug 2023
Moved: John D'Errico on 28 Aug 2023

Thank you al, Mr Nathan Hardenberg, Mr Dyuman Joshi and Mr Steven Lord,Â  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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### Accepted Answer

Nathan Hardenberg on 23 Aug 2023
Edited: Nathan Hardenberg on 23 Aug 2023
Here are some ideas, but please still regard my comment.
1. You have two variables x and p. You can not approximate numerically then.
2. "To approximate integrals directly, use vpaintegral instead of vpa. The vpaintegral function is faster and provides control over integration tolerances." [source]
3. Specify that you want to use x as your variable (int(fun, x, [a b]))
4. 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)))
funÂ =Â
vpa(int(fun, 0.001, 0.999)) % choosing values slightly above 0 and below 1
ansÂ =Â
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
ansÂ =Â
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'.
##### 8 CommentsShow 6 older commentsHide 6 older comments
Lahcen on 25 Aug 2023
Edited: Lahcen on 26 Aug 2023
Attached you will find a reference about A, concerning your question, (but they are references in French).
But using what you found like, one can find what my published. Indeed, Let a<1 and let put
by setting we find,
then
Now, this integral is converge if and only if
which is equivalent .
I hope you like the proof.
Nathan Hardenberg on 27 Aug 2023
@Lahcen Thanks a lot :D

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