Calculate double integrate of sin

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Ana Laura
Ana Laura am 31 Jan. 2024
Kommentiert: Torsten am 31 Jan. 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 Kommentar
Sam Chak
Sam Chak am 31 Jan. 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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Antworten (1)

Dyuman Joshi
Dyuman Joshi am 31 Jan. 2024
Bearbeitet: Dyuman Joshi am 31 Jan. 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 Kommentare
Ana Laura
Ana Laura am 31 Jan. 2024
Thanks very much. It's perfect!!
Torsten
Torsten am 31 Jan. 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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