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I am trying to do the following integral numerically,

$$\rho(\theta_{j},\phi_{j})=\int\frac{\sin{\theta_{i}} d\theta_{i} d\phi_{i}}{\sqrt{2+2[\cos(\theta_{i})\cos(\theta_{j})+\sin{\theta_{i}\sin{\theta_{j}\cos{(\phi_{i}-\phi_{j})]}}}}}$$

How do I do this in Mathematica? When I try, I get some error saying that the denominator becomes zeros. How do I avoid this issue of denominator becoming zeros?

Edit

Yes, the variables are on the surface of a sphere. The denominator is in fact dot product of two arbitrary vectors

$$\sqrt{2+2 \vec{d_{i}} . \vec{d_{j}}}$$

Here $\vec{d_{i}}$ and $\vec{d_{j}}$ are unit vectors on a sphere. The actual integral is

$$\int \frac{{d^{3}\vec{d_{i}}}}{\sqrt{2+2 \vec{d_{i}} . \vec{d_{j}}}}$$.

As those two vectors can be anywhere on the surface on the sphere, I wanted the integrand which is some

$$f(\vec{d_{i}},\vec{d_{j}})$$

to be integrated over one of the vetor

$$\int f(\vec{d_{i}},\vec{d_{j}}) d^{3}\vec{d_{i}}$$

to get some function which is a function of $$g(\vec{d_{j}})$$

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  • $\begingroup$ People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this meta Q&A helpful $\endgroup$
    – Michael E2
    Commented Apr 30, 2019 at 10:46
  • $\begingroup$ I cannot reproduce the Power:infy (divide by zero) error and have no trouble obtaining the correct result. Problems with code usually require the code to solve the problem. Please post your code. $\endgroup$
    – Michael E2
    Commented Apr 30, 2019 at 13:13

1 Answer 1

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The integral you are calculating is not a function of $\theta_j$, $\phi_j$. It’s a constant.

In order to see this, note that each pair of angles $(\theta,\phi)$ identifies a point on the surface of a sphere, and that the integrand (remembering that the numerator is the measure on a spherical surface) is a function of the angle between these two points.

By spherical invariance you can then calculate the integral for any value of $\theta_j$, $\phi_j$ you like, the result will be the same.

Set them to zero, the integral will simplify a lot, probably Mathematica can calculate it analytically.

Edit: looks like the integral is $4 \pi$.

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  • $\begingroup$ You are right the variables are on the surface of a sphere. The denominator is in fact dot product of two arbitrary vectors $$\sqrt{2+2 \vec{d_{i}} . \vec{d_{j}}}$$ Here $\vec{d_{i}}$ and $\vec{d_{j}}$ are unit vectors on a sphere. The actual integral is $$\int \frac{{d^{3}\vec{d_{i}}}}{\sqrt{2+2 \vec{d_{i}} . \vec{d_{j}}}}$$ $\endgroup$
    – user135580
    Commented Apr 30, 2019 at 7:18
  • $\begingroup$ I have edited the question. This is a vector function with 4 variables. Is it possible to get a vector function of two variables by integrating the other two variables $\endgroup$
    – user135580
    Commented Apr 30, 2019 at 7:27
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    $\begingroup$ Sure. But there's more to it. Your function $\int f(\vec{d_{i}},\vec{d_{j}}) d^{3}\vec{d_{i}}$ by rotational invariance is equivalent to $\int f(\vec{d_{i}}, \vec{0}) d^{3}\vec{d_{i}}$, which is just a number, does not depend on $\vec{d_{j}}$. $\endgroup$
    – zakk
    Commented Apr 30, 2019 at 11:58
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    $\begingroup$ Do what you say, the integral reduces to a first-year calculus integral $2\pi \int_0^\pi {\sin \theta \;d\theta \over \sqrt{2+2\cos\theta}}\,,$ and you probably don't even need Mathematica to calculate it symbolically :) $\endgroup$
    – Michael E2
    Commented Apr 30, 2019 at 13:10

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