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Proof of a definite multiple integral relation

  1. Feb 7, 2010 #1
    Hello,

    The following is identity no. 4.624 in Gradshteyn & Ryzhik's Table of Integrals, Series and Products:
    [tex]
    \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}
    [/tex]
    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 [tex]t=\cos p[/tex] will get you from the second to third integral.

    Thanks.
     
  2. jcsd
  3. Feb 7, 2010 #2

    mathman

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    Science Advisor

    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).
     
  4. Feb 8, 2010 #3
    Mathman,

    Thanks for your help. I have followed your advice on first [tex]\phi[/tex] and then [tex]\theta[/tex] and got the following.

    Starting from
    [tex]
    \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}
    [/tex]

    However, I cannot get beyond this point to reduce the term in the square root to [tex]R[/tex]. Am I missing something?
     
  5. Feb 9, 2010 #4

    mathman

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    Science Advisor

    The only thing I can suggest is that you try combining α with βcosφ + γsinφ first. Good luck!
     
    Last edited: Feb 9, 2010
  6. Feb 10, 2010 #5

    mathman

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    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!!!!!!!!
     
  7. Feb 11, 2010 #6
    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!
     
  8. Feb 15, 2010 #7
    Hi Mathman,

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

    [tex]\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} [/tex]

    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
    [tex]
    \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}
    [/tex]
    where
    [tex]
    \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}
    [/tex]
    with
    [tex]
    \begin{equation}
    R_1 = \sqrt{\alpha^2+\beta^2+\gamma^2},R_2 = \sqrt{a^2+b^2+c^2}\nonumber
    \end{equation}
    [/tex]
    I then rotated the coordinate system such that the z-axis (from which [tex]\theta[/tex] is measured) lies along the vector [tex]\mathbf{v_1}[/tex] which allows me to reduce the integral to
    [tex]
    \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}
    [/tex]
    where
    [tex]
    \begin{equation}
    \lambda = \arccos\left(\mathbf{v_1}.\mathbf{v_2}\right).\nonumber
    \end{equation}
    [/tex]

    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 [tex]\mathbf{v_1}[/tex] and [tex]\mathbf{v_2}[/tex] are parallel. Thanks in advance!
     
  9. Feb 15, 2010 #8

    mathman

    User Avatar
    Science Advisor

    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.
     
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