# Proof of a definite multiple integral relation

Hello,

The following is identity no. 4.624 in Gradshteyn & Ryzhik's Table of Integrals, Series and Products:
$$\int_0^{\pi}\int_0^{2\pi}f\left(\alpha\cos\theta+\beta\sin\theta\cos\phi+\gamma\sin\theta\sin\phi\right)\sin\theta d\theta d\phi=2\pi\int_0^{\pi}f\left(R\cos p\right)\sin p\, dp = 2\pi\int_{-1}^{1}f\left(Rt\right)dt\nonumber where R = \sqrt{\alpha^2+\beta^2+\gamma^2}.\nonumber$$
I would like to know how this can be proved, in particular, how to get from the first to the second integral. I know that a substitution of $$t=\cos p$$ will get you from the second to third integral.

Thanks.

mathman
I haven't worked out the details (I'll leave that to you),
but the general idea is repeated application of acosx + bsinx=r(cosxcosy+sinxsiny)=rcos(x-y)
where r² = a² + b² and y=arctan(b/a). You can also use an analogous trick to end up with sin(x+y).

Mathman,

Thanks for your help. I have followed your advice on first $$\phi$$ and then $$\theta$$ and got the following.

Starting from
$$\alpha\cos\theta+\beta\cos\phi\sin\theta+\gamma\sin\theta\sin\phi=\alpha\cos\theta+\sin\theta\left(\beta\cos\phi+\gamma\sin\phi\right)\nonumber which becomes \alpha\cos\theta+\sin\theta\left(\sqrt{\gamma^{2}+\beta^{2}}\cos\left(\phi-\arctan\frac{\gamma}{\beta}\right)\right)\nonumber and then \sqrt{\alpha^{2}+\left(\gamma^2+\beta^2\right)\cos^2\left(\phi-\arctan\frac{\gamma}{\beta}\right)}\cos\left(\theta-\arctan\left(\frac{\sqrt{\gamma^2+\beta^2}\cos\left(\phi-\arctan\frac{\gamma}{\beta}\right)}{\alpha}\right)\right)\nonumber$$

However, I cannot get beyond this point to reduce the term in the square root to $$R$$. Am I missing something?

mathman
The only thing I can suggest is that you try combining α with βcosφ + γsinφ first. Good luck!

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mathman
Ignore previous answers - I finally got it!! It can't be done one angle at a time, as far as I can tell. However looking at from a spherical coordinate point of view it is relatively simple (after much thought).

Start be defining a unit vector V = (α/R,β/R,γ/R), where R is as defined in the original problem.
Let Q=(x,y,z) be a point (in rectangular coordinates) on the surface of the sphere of radius R.
Then the argument of the function (f) is simply the dot product of V and Q.
This is independent of the angular coordinate system you are using, i.e. what you have is the integral of f(V.Q) over the entire surface of the sphere.
Choose a set of spherical coordinates with the axis in the same direction as V and you will get the second integral (or the equivalent with sin and cos switched and end points adjusted accordingly).

This will work!!!!!!!!

Thanks for your help Mathman, it makes sense to me now! I had begun to work on it from the point of view of the surface area of a sphere but hadn't managed to get it.

Thank you!

Hi Mathman,

I am trying to extend the method you suggested to solving integrals of the form

$$\int_0^{\pi}\int_0^{2\pi}\sin\theta f\left(\alpha\sin\theta\cos\phi+\beta\sin\theta\sin\phi+\gamma\cos\theta\right)g\left(a\sin\theta\cos\phi+b\sin\theta\sin\phi+c\cos\theta\right) d\theta d\phi\nonumber$$

As before, I have treated it as the dot product of a unit vector and a vector on a sphere. In this case there are two such dot products
$$\int_0^{\pi}\int_0^{2\pi}\sin\theta f\left(\mathbf{v_1}.\mathbf{Q_1}\right)g\left(\mathbf{v_2}.\mathbf{Q_2}\right) d\theta d\phi\nonumber$$
where
$$\mathbf{v_1}= \left(\frac{\alpha}{R_1},\frac{\beta}{R_1},\frac{\gamma}{R_1}\right), \mathbf{v_2}= \left(\frac{a}{R_2},\frac{b}{R_2},\frac{c}{R_2}\right)\nonumber$$
with
$$R_1 = \sqrt{\alpha^2+\beta^2+\gamma^2},R_2 = \sqrt{a^2+b^2+c^2}\nonumber$$
I then rotated the coordinate system such that the z-axis (from which $$\theta$$ is measured) lies along the vector $$\mathbf{v_1}$$ which allows me to reduce the integral to
$$\int_0^{\pi}\int_0^{2\pi}\sin \psi f\left(R_1\cos \psi\right)g\left(R_2\cos\left(\lambda-\psi\right)\cos\phi\right)d\phi d\psi\nonumber$$
where
$$\lambda = \arccos\left(\mathbf{v_1}.\mathbf{v_2}\right).\nonumber$$

I am wondering if there is anyway that I can reduce this to a single integral as was done in the previous case.So far, I have only managed to do this if $$\mathbf{v_1}$$ and $$\mathbf{v_2}$$ are parallel. Thanks in advance!

mathman