I am trying to do a six dimensional integration involving functions of two vectors $\vec{p}$ and $\vec{p}'$ in spherical coordinates. So $\vec{p}$ is reprsented by $\{p,\theta,\phi\}$, and $\vec{p}'$ is represented by $\{p',\theta',\phi'\}$. The $\theta$'s range from 0 to $\pi$, and the $\phi$'s range from 0 to $2\pi$. The angle between the two vectors is given by $\cos\gamma = \cos\theta\cos\theta' + \sin\theta\sin\theta'\cos(\phi-\phi')$. I cannot take any of these to vectors to be the $z$-axis, since the $z$-axis is fixed by a third vector.

Suppose the function that I have to integrate is $$ \frac{\cos(\phi-\phi')}{[p^2 + p'^2 + 2pp'\cos\gamma + M^2]^2} $$ where $M$ is a real constant. Then the integrals over $\phi$ and $\phi'$ would be of the form $$ \int_0^{2\pi} d\phi' \int_0^{2\pi} d\phi \frac{\cos(\phi-\phi')}{[a + b\cos(\phi - \phi')]^2} $$ where I have used $a$ and $b$ to represent the terms that are independent of $\phi$ and $\phi'$. If I now do the following substitution $$ \sin(\phi - \phi') = u $$ then the integration becomes $$ \int_0^{2\pi} d\phi' \int_{-\sin\phi'}^{-\sin\phi'} du\frac{1}{[a+b\sqrt{1-u^2}]^2} $$ which is just zero.

There are a bunch of other functions which involve $\cos(\phi-\phi')$ that I am trying to integrate. I was expecting most of these to be nonzero. But from what I see it appears to me that if it involves $\cos(\phi-\phi')$ then for the given range of $\phi$ the integral will always be zero.

Am I doing something wrong here?


The basic problem is that your substitution is not one to one in the entire range of the integral.

You can correct the situation by splitting up the integral using intervals on which it is one to one

note that $ \cos \theta = -\sqrt{1-\sin^2\theta}$ when $ \theta \in (\frac \pi 2, \frac {3\pi} 2)$

In General $$ I= \int_0^{2\pi}\cos\theta f(\cos \theta)d\theta \\= \int_0^{\frac \pi 2}\cos\theta f(\cos \theta)d\theta +\int_{\frac \pi 2}^{\frac {3\pi }2}\cos\theta f(\cos \theta)d\theta + \int_{\frac {3\pi }2}^{2\pi}\cos\theta f(\cos \theta)d\theta $$ now we can make the sine substitution in all three integrals, being careful to use $\cos\theta = -\sqrt{1-u^2} $ in the second integral $$ I=\int _0^1f(\sqrt{1-u^2})du+\int _{1}^{-1}f(-\sqrt{1-u^2})du +\int _{-1}^0f(\sqrt{1-u^2})du \\=\int_{-1}^1\Bigg[ f(\sqrt{1-u^2})-f(-\sqrt{1-u^2}) \Bigg]du $$


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