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Math Help - Approximating a difficult (or impossible?) integral

  1. #1
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    Question Approximating a difficult (or impossible?) integral

    Hello!

    I'd like to create a probability density function where the probability of observing a random variable is proportional to the distance from a circle.

    Let the pair (r, \theta) represent any location in polar coordinates and
    let the pair (r_1, \theta_1) be the centre of the circle of radius R.

    The distance between any point and the centre of the circle can be found using the law of cosines: d = \sqrt{r^2 + r_1^2 -2r r_1\cos(\theta - \theta_1)}

    Which means the distance to the closest point on the circle is D = |d - R|

    Now the distribution is Gaussian shaped so I set the pdf as:
    p(r, \theta) = e^{(-\frac{D^2}{2 \sigma^2})}

    The probability of a random variable falling in a polar area P_A is found by integration:

    P_A = \int_{r_a}^{r_b}\int_{\theta_a}^{\theta_b}p(r, \theta)rdr d \theta = \int_{r_a}^{r_b}\int_{\theta_a}^{\theta_b} e^{-\frac{(\sqrt{r^2 + r_1^2 - 2r r_1 \cos(\theta - \theta_1)} - R)^2}{2 \sigma^2}}  rdr d \theta

    But neither Maxima nor Mathematica were able to solve that integral.

    I started a Taylor expansion about a fixed point but it got ugly fast.

    Any suggestions for another method of solving or approximating this integral are appreciated!
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  2. #2
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    Quote Originally Posted by jabernathy View Post
    Hello!

    I'd like to create a probability density function where the probability of observing a random variable is proportional to the distance from a circle.

    Let the pair (r, \theta) represent any location in polar coordinates and
    let the pair (r_1, \theta_1) be the centre of the circle of radius R.

    The distance between any point and the centre of the circle can be found using the law of cosines: d = \sqrt{r^2 + r_1^2 -2r r_1\cos(\theta - \theta_1)}

    Which means the distance to the closest point on the circle is D = |d - R|

    Now the distribution is Gaussian shaped so I set the pdf as:
    p(r, \theta) = e^{(-\frac{D^2}{2 \sigma^2})}

    The probability of a random variable falling in a polar area P_A is found by integration:

    P_A = \int_{r_a}^{r_b}\int_{\theta_a}^{\theta_b}p(r, \theta)rdr d \theta = \int_{r_a}^{r_b}\int_{\theta_a}^{\theta_b} e^{-\frac{(\sqrt{r^2 + r_1^2 - 2r r_1 \cos(\theta - \theta_1)} - R)^2}{2 \sigma^2}}  rdr d \theta

    But neither Maxima nor Mathematica were able to solve that integral.

    I started a Taylor expansion about a fixed point but it got ugly fast.

    Any suggestions for another method of solving or approximating this integral are appreciated!
    I haven't tried this question out myself yet, but I do remember Rule Number 6... which is reverse the order of integration...
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  3. #3
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    Changing the order of integration didn't improve the results.

    Maxima yielded this answer:

    \int_{a}^{b}x\ \int_{\alpha}^{\beta}{e}^{-\frac{{\left( \sqrt{{y}^{2}-2\,cos\left( \theta-\phi\right) x y+{x}^{2}}-z\right) }^{2}}{2\,{\sigma}^{2}}}d\theta dx

    and from earlier:

    \int_{\alpha}^{\beta}\int_{a}^{b}x\,{e}^{-\frac{{\left( \sqrt{{y}^{2}-2\ cos\left( \theta-\phi\right) x y+{x}^{2}}-z\right) }^{2}}{2\,{\sigma}^{2}}}dxd\theta

    (I had to change the variables because Maxima doesn't like arguments with subscripts IIRC).

    Is a series expansion the next logical step?
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