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I have ended up to this integral ##\int_{φ=0}^{2π} \int_{θ=0}^π \sin θ \ \cos θ~Y_{00}^*~Y_{00}~dθ \, dφ## while I was solving a problem.

I know that in spherical coordinates when ##\vec r → -\vec r## :

1) The magnitude of ##\vec r## does not change : ##r' → r##

2) The angles ##θ## and ##φ## change like ##θ' → π-θ## and ##φ' → π+φ##

3) So parity of spherical harmonics is ##\hat P## ##Y_{lm}(θ,φ)## ##=## ##Y_{lm}(θ',φ')=(-1)^l## ##Y_{lm}(θ,φ)##

4) Parity of ##\cos θ## and ##\sin θ## are ##\cos θ'## ##=## ##(-1)## ##\cos θ## and ##\sin θ'## ##=## ##\sin θ## respectively.

This means that in my case the integrated function has Parity equal to ##(-1)## .

So my question is:

Can I say that this integral is zero because of the odd (=parity is equal to (-1)) integrated funtion? Because in the xy plane when we integrate an odd function ##F(x)=-F(-x)## in a symmetric space (e.g. ##\int_{-a}^a F(x) \, dx## with F being odd) we can say that it is zero without to calculate it.

If yes, can I do this in general? meaning, if i get a random function in spherical coordinates which depends only from angles θ and φ and i want to integrate it with these limits: ##\int_{φ=0}^{2π} \int_{θ=0}^π randomF(θ,φ) \, dθ \, dφ## , can i find its parity and say if it is zero or not?

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# Parity and integration in spherical coordinates

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