Integrating multiple products of order 1 Legendre functions
I would like a closed form expression for the integrals of some products like the following:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
$$
and
$$
int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
$$
where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):
$$
P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
$$
where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.
Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!
integration legendre-polynomials legendre-functions
asked Dec 8 at 16:45
Pan Daemonium
526
add a comment |
I would like a closed form expression for the integrals of some products like the following:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
$$
and
$$
int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
$$
where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):
$$
P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
$$
where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.
Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!
integration legendre-polynomials legendre-functions
asked Dec 8 at 16:45
Pan Daemonium
526
add a comment |
I would like a closed form expression for the integrals of some products like the following:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
$$
and
$$
int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
$$
where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):
$$
P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
$$
where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.
Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!
integration legendre-polynomials legendre-functions
asked Dec 8 at 16:45
Pan Daemonium
526
I would like a closed form expression for the integrals of some products like the following:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
$$
and
$$
int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
$$
where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):
$$
P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
$$
where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.
Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!
integration legendre-polynomials legendre-functions
integration legendre-polynomials legendre-functions
asked Dec 8 at 16:45
Pan Daemonium
526
asked Dec 8 at 16:45
Pan Daemonium
526
edited Dec 8 at 23:10
asked Dec 8 at 16:45
Pan Daemonium
526
asked Dec 8 at 16:45
Pan Daemonium
526
asked Dec 8 at 16:45
Pan Daemonium
526
526
add a comment |
add a comment |
1 Answer
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I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
$$
where I've defined
$$
G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
$$
in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.
The other integral can be written out similarly.
(Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)
answered Dec 8 at 23:27
Pan Daemonium
526
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
$$
where I've defined
$$
G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
$$
in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.
The other integral can be written out similarly.
(Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)
answered Dec 8 at 23:27
Pan Daemonium
526
add a comment |
I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
$$
where I've defined
$$
G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
$$
in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.
The other integral can be written out similarly.
(Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)
answered Dec 8 at 23:27
Pan Daemonium
526
add a comment |
I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
$$
where I've defined
$$
G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
$$
in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.
The other integral can be written out similarly.
(Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)
answered Dec 8 at 23:27
Pan Daemonium
526
I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:
$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
$$
where I've defined
$$
G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
$$
in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.
The other integral can be written out similarly.
(Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)
answered Dec 8 at 23:27
Pan Daemonium
526
edited Dec 8 at 23:33
answered Dec 8 at 23:27
Pan Daemonium
526
answered Dec 8 at 23:27
Pan Daemonium
526
answered Dec 8 at 23:27
Pan Daemonium
526
526
add a comment |
add a comment |
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