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
edited Nov 23 at 8:13
user10354138
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
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up vote
1
down vote
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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
edited Nov 23 at 8:13
user10354138
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
264
add a comment |
up vote
1
down vote
favorite
up vote
1
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
edited Nov 23 at 8:13
user10354138
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
264
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
edited Nov 23 at 8:13
user10354138
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
264
edited Nov 23 at 8:13
user10354138
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
264
edited Nov 23 at 8:13
user10354138
6,9951624
edited Nov 23 at 8:13
user10354138
6,9951624
edited Nov 23 at 8:13
user10354138
6,9951624
6,9951624
asked Nov 22 at 15:08
Dimension Obscura
264
asked Nov 22 at 15:08
Dimension Obscura
264
asked Nov 22 at 15:08
Dimension Obscura
264
264
add a comment |
add a comment |
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