Integration of product of Associated Legendre Polynomial
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I am interested in the following integral $$I=int_{-1}^{1} P_s^t(x)P_u^v(x)mathrm{d}x~.$$ Does any one know if a closed form exist for a general $s, t, u, v$, and for $tneq v$ and $sneq u$?
integration definite-integrals legendre-polynomials
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
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add a comment |
$begingroup$
I am interested in the following integral $$I=int_{-1}^{1} P_s^t(x)P_u^v(x)mathrm{d}x~.$$ Does any one know if a closed form exist for a general $s, t, u, v$, and for $tneq v$ and $sneq u$?
integration definite-integrals legendre-polynomials
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
$endgroup$
$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34
add a comment |
$begingroup$
I am interested in the following integral $$I=int_{-1}^{1} P_s^t(x)P_u^v(x)mathrm{d}x~.$$ Does any one know if a closed form exist for a general $s, t, u, v$, and for $tneq v$ and $sneq u$?
integration definite-integrals legendre-polynomials
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
$endgroup$
I am interested in the following integral $$I=int_{-1}^{1} P_s^t(x)P_u^v(x)mathrm{d}x~.$$ Does any one know if a closed form exist for a general $s, t, u, v$, and for $tneq v$ and $sneq u$?
integration definite-integrals legendre-polynomials
integration definite-integrals legendre-polynomials
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
asked Sep 16 '16 at 23:08
titaniumtitanium
320111
320111
$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34
add a comment |
$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34
$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34
$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34
add a comment |
1 Answer
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I think your answer is in the paper, "The overlap integral of three associated Legendre polynomials" by Shi-Hai Dong and R. Lemus (2002) (https://www.sciencedirect.com/science/article/pii/S0893965902800040). The relevant expression is $I(s,t;u,v)$, given by equations (7-10), and yields
$$
int_{-1}^1 P^t_s(x) P^v_u(x)dx=(-1)^delta|v-t|2^{|v-t|-2}sqrt{frac{(s+t)!(u+v)!}{(s-t)!(u-v)!}}timessum_alpha (2alpha+1)left(1+(-1)^{alpha+|v-t|}right)sqrt{frac{(alpha-|v-t|)!}{(alpha+|v-t|)!}}frac{Gammaleft(frac{alpha}{2}right)Gammaleft(frac{alpha+|v-t|+1}{2}right)}{left(frac{alpha-|v-t|}{2}right)!Gammaleft(frac{alpha+3}{2}right)},
$$
with the phase
$$
delta=left{begin{matrix} m_1 & mbox{ if } & m_2geq m_1, \ m_2 & mbox{ if } & m_2<m_1,end{matrix}right.
$$
and this integral will only be non-zero if
$$
(|s-u|leqalphaleq s+u)wedge(alphageq|v-t|)wedge(alpha + s + umbox{ is even}).
$$
Note that there is nothing specifying that $tneq v$ or $sneq u$, so this formula may simplify a little under those assumption, I haven't checked.
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think your answer is in the paper, "The overlap integral of three associated Legendre polynomials" by Shi-Hai Dong and R. Lemus (2002) (https://www.sciencedirect.com/science/article/pii/S0893965902800040). The relevant expression is $I(s,t;u,v)$, given by equations (7-10), and yields
$$
int_{-1}^1 P^t_s(x) P^v_u(x)dx=(-1)^delta|v-t|2^{|v-t|-2}sqrt{frac{(s+t)!(u+v)!}{(s-t)!(u-v)!}}timessum_alpha (2alpha+1)left(1+(-1)^{alpha+|v-t|}right)sqrt{frac{(alpha-|v-t|)!}{(alpha+|v-t|)!}}frac{Gammaleft(frac{alpha}{2}right)Gammaleft(frac{alpha+|v-t|+1}{2}right)}{left(frac{alpha-|v-t|}{2}right)!Gammaleft(frac{alpha+3}{2}right)},
$$
with the phase
$$
delta=left{begin{matrix} m_1 & mbox{ if } & m_2geq m_1, \ m_2 & mbox{ if } & m_2<m_1,end{matrix}right.
$$
and this integral will only be non-zero if
$$
(|s-u|leqalphaleq s+u)wedge(alphageq|v-t|)wedge(alpha + s + umbox{ is even}).
$$
Note that there is nothing specifying that $tneq v$ or $sneq u$, so this formula may simplify a little under those assumption, I haven't checked.
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
$endgroup$
add a comment |
$begingroup$
I think your answer is in the paper, "The overlap integral of three associated Legendre polynomials" by Shi-Hai Dong and R. Lemus (2002) (https://www.sciencedirect.com/science/article/pii/S0893965902800040). The relevant expression is $I(s,t;u,v)$, given by equations (7-10), and yields
$$
int_{-1}^1 P^t_s(x) P^v_u(x)dx=(-1)^delta|v-t|2^{|v-t|-2}sqrt{frac{(s+t)!(u+v)!}{(s-t)!(u-v)!}}timessum_alpha (2alpha+1)left(1+(-1)^{alpha+|v-t|}right)sqrt{frac{(alpha-|v-t|)!}{(alpha+|v-t|)!}}frac{Gammaleft(frac{alpha}{2}right)Gammaleft(frac{alpha+|v-t|+1}{2}right)}{left(frac{alpha-|v-t|}{2}right)!Gammaleft(frac{alpha+3}{2}right)},
$$
with the phase
$$
delta=left{begin{matrix} m_1 & mbox{ if } & m_2geq m_1, \ m_2 & mbox{ if } & m_2<m_1,end{matrix}right.
$$
and this integral will only be non-zero if
$$
(|s-u|leqalphaleq s+u)wedge(alphageq|v-t|)wedge(alpha + s + umbox{ is even}).
$$
Note that there is nothing specifying that $tneq v$ or $sneq u$, so this formula may simplify a little under those assumption, I haven't checked.
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
$endgroup$
add a comment |
$begingroup$
I think your answer is in the paper, "The overlap integral of three associated Legendre polynomials" by Shi-Hai Dong and R. Lemus (2002) (https://www.sciencedirect.com/science/article/pii/S0893965902800040). The relevant expression is $I(s,t;u,v)$, given by equations (7-10), and yields
$$
int_{-1}^1 P^t_s(x) P^v_u(x)dx=(-1)^delta|v-t|2^{|v-t|-2}sqrt{frac{(s+t)!(u+v)!}{(s-t)!(u-v)!}}timessum_alpha (2alpha+1)left(1+(-1)^{alpha+|v-t|}right)sqrt{frac{(alpha-|v-t|)!}{(alpha+|v-t|)!}}frac{Gammaleft(frac{alpha}{2}right)Gammaleft(frac{alpha+|v-t|+1}{2}right)}{left(frac{alpha-|v-t|}{2}right)!Gammaleft(frac{alpha+3}{2}right)},
$$
with the phase
$$
delta=left{begin{matrix} m_1 & mbox{ if } & m_2geq m_1, \ m_2 & mbox{ if } & m_2<m_1,end{matrix}right.
$$
and this integral will only be non-zero if
$$
(|s-u|leqalphaleq s+u)wedge(alphageq|v-t|)wedge(alpha + s + umbox{ is even}).
$$
Note that there is nothing specifying that $tneq v$ or $sneq u$, so this formula may simplify a little under those assumption, I haven't checked.
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
$endgroup$
I think your answer is in the paper, "The overlap integral of three associated Legendre polynomials" by Shi-Hai Dong and R. Lemus (2002) (https://www.sciencedirect.com/science/article/pii/S0893965902800040). The relevant expression is $I(s,t;u,v)$, given by equations (7-10), and yields
$$
int_{-1}^1 P^t_s(x) P^v_u(x)dx=(-1)^delta|v-t|2^{|v-t|-2}sqrt{frac{(s+t)!(u+v)!}{(s-t)!(u-v)!}}timessum_alpha (2alpha+1)left(1+(-1)^{alpha+|v-t|}right)sqrt{frac{(alpha-|v-t|)!}{(alpha+|v-t|)!}}frac{Gammaleft(frac{alpha}{2}right)Gammaleft(frac{alpha+|v-t|+1}{2}right)}{left(frac{alpha-|v-t|}{2}right)!Gammaleft(frac{alpha+3}{2}right)},
$$
with the phase
$$
delta=left{begin{matrix} m_1 & mbox{ if } & m_2geq m_1, \ m_2 & mbox{ if } & m_2<m_1,end{matrix}right.
$$
and this integral will only be non-zero if
$$
(|s-u|leqalphaleq s+u)wedge(alphageq|v-t|)wedge(alpha + s + umbox{ is even}).
$$
Note that there is nothing specifying that $tneq v$ or $sneq u$, so this formula may simplify a little under those assumption, I haven't checked.
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
answered Dec 9 '18 at 0:03
Pan DaemoniumPan Daemonium
526
526
add a comment |
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$begingroup$
Legendre and Related Functions .
$endgroup$
– Felix Marin
Sep 17 '16 at 2:34