Evaluating an integral using Gegenbauer polynomials











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I want to evaluate the following integral
$$int frac{(r-r'costheta')^2r'^2,dr'sintheta',dtheta',dphi'}{(r^2+r'^2-2rr'costheta')^{3/2}}$$
Working that a little bit i end up with this expression.
$$frac{2pi}{r^3} int_0^pi dtheta'frac{r^2 (1-frac{r'}{r}costheta')^2sintheta'}{(1+(frac{r'}{r})^2-2frac{r'}{r}costheta' )^{3/2}}int_0^R dr',r'^2$$
Using Gegenbauer polynomials to express the denominator i get
$ frac{2pi}{r} int_0^pi dtheta'(1-frac{r'}{r}costheta')^2sintheta'sum C_n(costheta')(frac{r'}{r})^n int_0^R dr'r'^2$ Doing some more calculations i get the following integrals.
$int_{-1}^1dx sum C_n(x)$ $int_{-1}^1dx sum C_n(x)x^2$ $int_{-1}^1dx sum C_n(x)x$

Which i dont know how to evaluate since I'm missing the weight function $(1-x^2)$ in the integrals






orthogonal-polynomials







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    I want to evaluate the following integral
    $$int frac{(r-r'costheta')^2r'^2,dr'sintheta',dtheta',dphi'}{(r^2+r'^2-2rr'costheta')^{3/2}}$$
    Working that a little bit i end up with this expression.
    $$frac{2pi}{r^3} int_0^pi dtheta'frac{r^2 (1-frac{r'}{r}costheta')^2sintheta'}{(1+(frac{r'}{r})^2-2frac{r'}{r}costheta' )^{3/2}}int_0^R dr',r'^2$$
    Using Gegenbauer polynomials to express the denominator i get
    $ frac{2pi}{r} int_0^pi dtheta'(1-frac{r'}{r}costheta')^2sintheta'sum C_n(costheta')(frac{r'}{r})^n int_0^R dr'r'^2$ Doing some more calculations i get the following integrals.
    $int_{-1}^1dx sum C_n(x)$ $int_{-1}^1dx sum C_n(x)x^2$ $int_{-1}^1dx sum C_n(x)x$

    Which i dont know how to evaluate since I'm missing the weight function $(1-x^2)$ in the integrals






    orthogonal-polynomials







    share|cite|improve this question


























      up vote
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      up vote
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      down vote

      favorite











      I want to evaluate the following integral
      $$int frac{(r-r'costheta')^2r'^2,dr'sintheta',dtheta',dphi'}{(r^2+r'^2-2rr'costheta')^{3/2}}$$
      Working that a little bit i end up with this expression.
      $$frac{2pi}{r^3} int_0^pi dtheta'frac{r^2 (1-frac{r'}{r}costheta')^2sintheta'}{(1+(frac{r'}{r})^2-2frac{r'}{r}costheta' )^{3/2}}int_0^R dr',r'^2$$
      Using Gegenbauer polynomials to express the denominator i get
      $ frac{2pi}{r} int_0^pi dtheta'(1-frac{r'}{r}costheta')^2sintheta'sum C_n(costheta')(frac{r'}{r})^n int_0^R dr'r'^2$ Doing some more calculations i get the following integrals.
      $int_{-1}^1dx sum C_n(x)$ $int_{-1}^1dx sum C_n(x)x^2$ $int_{-1}^1dx sum C_n(x)x$

      Which i dont know how to evaluate since I'm missing the weight function $(1-x^2)$ in the integrals






      orthogonal-polynomials







      share|cite|improve this question















      I want to evaluate the following integral
      $$int frac{(r-r'costheta')^2r'^2,dr'sintheta',dtheta',dphi'}{(r^2+r'^2-2rr'costheta')^{3/2}}$$
      Working that a little bit i end up with this expression.
      $$frac{2pi}{r^3} int_0^pi dtheta'frac{r^2 (1-frac{r'}{r}costheta')^2sintheta'}{(1+(frac{r'}{r})^2-2frac{r'}{r}costheta' )^{3/2}}int_0^R dr',r'^2$$
      Using Gegenbauer polynomials to express the denominator i get
      $ frac{2pi}{r} int_0^pi dtheta'(1-frac{r'}{r}costheta')^2sintheta'sum C_n(costheta')(frac{r'}{r})^n int_0^R dr'r'^2$ Doing some more calculations i get the following integrals.
      $int_{-1}^1dx sum C_n(x)$ $int_{-1}^1dx sum C_n(x)x^2$ $int_{-1}^1dx sum C_n(x)x$

      Which i dont know how to evaluate since I'm missing the weight function $(1-x^2)$ in the integrals






      orthogonal-polynomials




      orthogonal-polynomials






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      edited Nov 23 at 8:13









      user10354138

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      asked Nov 22 at 15:08









      Dimension Obscura

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