matlab integration using function

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psychmohana ramesha
编辑： John D'Errico 2020-7-21
can i integrate this in matlab
integration of x^b/((1+x^b)^2) from 0 to infinity
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Walter Roberson
Walter Roberson 2020-7-21
integral() with anonymous function that uses vectorized operations such as ./ and .^

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John D'Errico
John D'Errico 2020-7-21
编辑：John D'Errico 2020-7-21
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.

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