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Calculate double integrate of sin

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Ana Laura
Ana Laura el 31 de En. de 2024
Comentada: Torsten el 31 de En. de 2024
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 comentario
Sam Chak
Sam Chak el 31 de En. de 2024
@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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Respuestas (1)

Dyuman Joshi
Dyuman Joshi el 31 de En. de 2024
Editada: Dyuman Joshi el 31 de En. de 2024
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 comentarios
Ana Laura
Ana Laura el 31 de En. de 2024
Thanks very much. It's perfect!!
Torsten
Torsten el 31 de En. de 2024
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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