Problem with using symbolic integration

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?

[Yasho Bharat Boggarapu]

sorry , I forgot to mention "syms x" in the first line.

Answer 1

[David Goodmanson]

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.

[Yasho Bharat Boggarapu]

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]

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.

[Yasho Bharat Boggarapu]

Thank you David. It was very useful.

[Yasho Bharat Boggarapu]

But ,I have noticed that beta calculates for any real value of (p,q). Then ,how does it affect the integration ?

[David Goodmanson]

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.