matlab integration using function

can i integrate this in matlab

integration of x^b/((1+x^b)^2) from 0 to infinity

[Walter Roberson]

integral() with anonymous function that uses vectorized operations such as ./ and .^

Answer 1

[John D'Errico]

Um, no. And, yes. Sort of. Is that a good, definiitive answer? Depending on the value of b, this has a solution, or no solution at all.

Pick some fixed value of b. Say b == 2.

b = 2;
syms x
int(x^b/(1+x^b)^2,0,inf)
ans =
pi/4

Larger values of b yield a little less clean solutions.

b = 3;
int(x^b/(1+x^b)^2,0,inf)
ans =
(2*pi*3^(1/2))/27

b = 4;
int(x^b/(1+x^b)^2,0,inf)
ans =
(2^(1/2)*(4*pi - log(- 1/2 - 1i/2)*(1 + 1i) - log(- 1/2 + 1i/2)*(1 - 1i) + log(1/2 - 1i/2)*(1 - 1i) + log(1/2 + 1i/2)*(1 + 1i)))/32

vpa(ans)
ans =
0.27768018363489789

As you can see, when b == 4, I had to resort to just computing the numerical value itself to give a clean, simple looking result.

But when b == 1,

b = 1;
int(x^b/(1+x^b)^2,0,inf)
ans =
Inf

Here we see a problem arises. With a little thought you can even see whay this fails when b == 1.

For general b>1, the integral is not some simple function of b. For b <= 1, a solution does not exist.

And if your question would be, when b > 1, is there some simply written algebraic function of b that provides a general solution, the answer seems to be no. You could define this as a special function, give it a name, tablulate values, etc. For b > 1, the problem is well-posed. But that is about as much as you could do.

So, can you form the integral? Well yes and no. Sort of.