Problem with using symbolic integration

2020 5 6
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更新时间：2020 5 6
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This is my expression
l = 0.0357;
u = 0.9;
p = 5.1;
q =3;
f= ((x-l)^(p-1)*(u-x)^(q-1)/(beta(p,q)*(u-l)^(p+q-1)));
F = (int(f,x,[l x]))
For these values it gives me a solution , where as when I change the value of say q=3.1 , it doesnt solve it . I get this back with some big integers
int((2305843009213693952*(9/10 - x)^(21/10)*(x - 357/10000)^(41/10))/6549555978944313, x, 357/10000, x)
Can anyone help me find the problem?

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David Goodmanson
David Goodmanson 2020-5-6

1 个投票

Hello Yasho,
For any reasonable q, integrating from l to u gives F = 1 by definition of the beta function. When q is an integer, integration from l to an arbitrary x0 gives an answer because symbolic can expand out (u-x)^(q-1) as a polynomial. So for q = 3 you can see a quadratic as part of the answer.
When both p and q are nonintegers and an arbitrary x0 as the upper limit, you have an incomplete beta function which has to be done numerically.

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Hi David,
Thanks for your reply .
It is a probability density function. So it will give 1 for numerical integration for any p,q(eventhough it is non integer).
syms x
l = 0.0357;
u = 0.9;
p = 4.9937;
q =2.5563;
f= ((x-l)^(p-1)*(u-x)^(q-1)/(beta(p,q)*(u-l)^(p+q-1)));
F = (vpaintegral(f,x,[l u]))
it gives F=1;
I am looking for an expression in x to solve further. It is just part of an expression.
I have given p as non integer, still it works right ? it is also part of beta(p,q). But , why it is just with q ?
David Goodmanson
David Goodmanson 2020-5-6
编辑：David Goodmanson 2020-5-6
Hi Yasho, for an arbitrary upper limit x0 it can do the integration if either p or q is an integer. But if neither p nor q are integers you are not going to get a result. Except numerically. And under the circumstances, there's nothing wrong with that.
Thank you David. It was very useful.
But ,I have noticed that beta calculates for any real value of (p,q). Then ,how does it affect the integration ?
David Goodmanson
David Goodmanson 2020-5-6
it allows you to always have an answer for the denominator of the integrand which is good, but it does not affect the conclusion in the previous comment I wrote.

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