Proof of a definite multiple integral relation

[appelberry]

Hello,

The following is identity no. 4.624 in Gradshteyn & Ryzhik's Table of Integrals, Series and Products:
$$\begin{equation} \int_0^{\pi}\int_0^{2\pi}f\left(\alpha\cos\theta+\beta\sin\theta\cos\phi+\gamma\sin\theta\sin\phi\right)\sin\theta d\theta d\phi=2\pi\int_0^{\pi}f\left(R\cos p\right)\sin p\, dp = 2\pi\int_{-1}^{1}f\left(Rt\right)dt\nonumber \end{equation} where \begin{equation} R = \sqrt{\alpha^2+\beta^2+\gamma^2}.\nonumber \end{equation}$$
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).

 

[appelberry]

Mathman,

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

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

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!

 

[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!

 

[appelberry]

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!

 

[appelberry]

Hi Mathman,

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

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

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
$$\begin{equation} \int_0^{\pi}\int_0^{2\pi}\sin\theta f\left(\mathbf{v_1}.\mathbf{Q_1}\right)g\left(\mathbf{v_2}.\mathbf{Q_2}\right) d\theta d\phi\nonumber \end{equation}$$
where
$$\begin{equation} \mathbf{v_1}= \left(\frac{\alpha}{R_1},\frac{\beta}{R_1},\frac{\gamma}{R_1}\right), \mathbf{v_2}= \left(\frac{a}{R_2},\frac{b}{R_2},\frac{c}{R_2}\right)\nonumber \end{equation}$$
with
$$\begin{equation} R_1 = \sqrt{\alpha^2+\beta^2+\gamma^2},R_2 = \sqrt{a^2+b^2+c^2}\nonumber \end{equation}$$
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
$$\begin{equation} \int_0^{\pi}\int_0^{2\pi}\sin \psi f\left(R_1\cos \psi\right)g\left(R_2\cos\left(\lambda-\psi\right)\cos\phi\right)d\phi d\psi\nonumber \end{equation}$$
where
$$\begin{equation} \lambda = \arccos\left(\mathbf{v_1}.\mathbf{v_2}\right).\nonumber \end{equation}$$

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]

After I posted the note for the previous solution I realized that it is slightly simpler if V = (α,β,γ) so V would have length R and the sphere would now have unit radius.

Generalizing to your second problem is better using this approach, since you would have v1 = (α,β,γ) abd v2=(a,b,c) allowing the same Q to be used in the arguments for f and g.
I don't know if this helps.