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Calculate double integrate of sin

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Ana Laura
Ana Laura 2024-1-31
评论： Torsten 2024-1-31
I want to calculate this:
But I can't write it correctly. I get the gamma function all the time.
It seems to be something like this:
syms theta rho n
f = sin(rho)^(n + 2)*sin(theta)^(n + 1)
I1 = int(f,rho,0,pi)
I2 = int(I1,theta,0,pi)
But the answer is:
(2*2^(2*n)*gamma(n/2)^2*gamma(n/2 + 1/2)^2*(n/2 + 1/2)^2)/(gamma(n)^2*(n + 1)^2*(n + 2))
When, i hope the return was:
2pi/n+2
Thanks in advance.
  1 个评论
Sam Chak
Sam Chak 2024-1-31
@Ana Laura, Erm... I believe the Gamma Γ symbols (not gamma γ) are the Gamma function. I don't use it often, but I encountered it in the Riemann zeta function.

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回答（1 个）

Dyuman Joshi
Dyuman Joshi 2024-1-31
编辑：Dyuman Joshi 2024-1-31
Ran in:
You can simplify the expression obtained -
syms theta rho n
f = sin(rho)^(n + 2)*sin(theta)^(n + 1);
I1 = int(f,rho,0,pi);
I2 = int(I1,theta,0,pi)
I2 = 
%Output
out = simplify(I2)
out = 
  2 个评论
Ana Laura
Ana Laura 2024-1-31
Thanks very much. It's perfect!!
Torsten
Torsten 2024-1-31
Ran in:
You can also write the double integral as the product of two one-dimensional integrals:
syms rho n
f = int(sin(rho)^(n + 2),rho,0,pi)*int(sin(rho)^(n + 1),rho,0,pi)
f = 
simplify(f)
ans = 

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