explicit integral could not be found

Hi,

I am trying to calculate some statistics using Matlab and I have almost completed it. All i need now is to calculate 1 number which will come from an integral. However the message Explicit integral cannot be found keeps coming up. here is my code:

syms x C;

double(int((x^1.7653)* ((1-x)^3.0546-1), x, C))

I have also used the same code but without the double() and still the same answer. Any help would be enormously appreciated. Thanks.

Answer 1

[Roger Stafford]

In the integrand, do you mean

   (x^1.7653)* ((1-x)^3.0546-1)      (Case 1)

or do you mean

   (x^1.7653)* (1-x)^(3.0546-1) ?    (Case 2)

In either case, matlab can give you the numerical value of the integral directly using its 'betacdf' function in the Statistics Toolbox. As you have seen, the 'int' function is unable to supply an explicit expression. That can easily happen.

It isn't clear what your integration limits are. I will suppose you want to integrate from x = 0 to x = c for some arbitrary parameter c with 0<=c<=1.

In Case 1 the integrand can be expressed as:

   x^1.7653*(1-x)^3.0546 - x^1.7653

and its integral would be

   gamma(a)*gamma(b)/gamma(a+b)*betacdf(c,a,b) - c^a/a

where a = 1.7653+1 and b = 3.0546+1.

In Case 2 its integral is:

   gamma(a)*gamma(b)/gamma(a+b)*betacdf(c,a,b)

where a = 1.7653+1 and b = 3.0546.

If your integration limits are different from the above, the above can easily be modified to give the correct answer. Just let us know what those limits actually are.

Answer 2

[Walter Roberson]

Are you intending to integrate over x = x to C ? Your form is a bit ambiguous about the limits.

Even without that you simply have the problem that it is difficult to find a closed form solution for the integral.

Perhaps it would be within your accuracy requirements to convert the exponents to rational, taylor that, and integrate the result?

Answer 3

[Peter]

It was case 1 Roger and what i was wanting to do was find a value for C cause I already knew the result of the integral. I found out how to do this though, it is an incomplete beta function which matalb has a function for.

betainc(X,W,Z)

where W and Z are the constants above. I then kept changing the values of X till i got the result I was looking for.

Thanks for all your time and effort.

[Roger Stafford]

The 'betainc' and 'betacdf' functions give the same result.

Do you agree with the term c^a/a to be subtracted in your integral?

You could have used 'fzero' to do the searching for the correct value for c.