# How to prove that the integral below in independent of $\theta$ and $\phi$?

Consider the following triple integral: $$\int_0^{2\pi}\int_0^\pi \int_0^1 \frac{1}{h} \frac{1-\alpha Q^2}{1 + \alpha Q^2 (1-\alpha Q^2)} \frac{\sin \vartheta}{\sqrt{1-Q^2}} \, \mathrm{d}Q \, \mathrm{d}\vartheta\, \mathrm{d} \varphi \, ,$$ where $$h^2 = 2a^2 \left( 1-\sin\theta \sin\vartheta \cos(\phi-\varphi)-\cos\theta\cos\vartheta \right) \, ,$$ and $$\alpha = \left(\frac{a}{h}\right)^2 \left[ \sin^2\theta+\sin^2\vartheta-2 \sin\theta \sin\vartheta \cos(\phi-\varphi) \right] \, .$$ Here, $a$ is a strictly positive real number (particle radius).

Actually, the integral quantifies a certain physical quantity that is supposed to be constant at the surface of the sphere. Therefore, the integral is expected to be independent of $\theta$ and $\phi$ which represent the polar and azimuthal angles, respectively. Numerically, this can be checked but I was wondering whether there exists analytical techniques that can be used to show that in a rigorous way.

Any help is highly appreciated.

Thanks

Hartmut Helmut

• @L.G. Q is an integration variable between 0 and 1. I will rewrite the integral – Math Student Sep 9 '17 at 18:31
• If you know what this integral means physically and how it is constructed does that not tell you why it is independent of the two variables? – Paul Sep 9 '17 at 18:36
• @Paul In fact, the integral is supposed to be so but still this needs to be proved. Thanks – Math Student Sep 9 '17 at 18:38
• No need for proof. $h,\alpha$ are constants. – Narasimham Sep 14 '17 at 4:13

Let $I$ be your integral. Since you are not integrating over $\theta, \phi$, the easiest way to show this is to show that
$$\frac{\partial I}{\partial \theta}=\frac{\partial I}{\partial \phi} = 0.$$
Since your variables of integration are $\vartheta, \varphi$ and $Q$, you can just pull both derivatives inside of the integral, so that you don't have to do any integration at all. If indeed your integral is independent of $\theta, \phi$, this should be quite simple.