The Associated Legendre Functions are regular solutions to the
general Legendre equation:
(
[
1
−
x
2
]
y
′
)
′
+
(
l
[
l
+
1
]
−
m
2
1
−
x
2
)
y
=
0
{\displaystyle \left(\left[1-x^{2}\right]y^{'}\right)^{'}+\left(l\left[l+1\right]-{\frac {m^{2}}{1-x^{2}}}\right)y=0}
z
′
=
d
z
d
x
.
{\displaystyle z^{'}={\frac {dz}{dx}}.}
This equation is an example of a more general class of equations
known as the Sturm-Liouville equations . Using Sturm-Liouville
theory, one can show that
K
k
l
m
=
∫
−
1
1
P
k
m
(
x
)
P
l
m
(
x
)
d
x
{\displaystyle K_{kl}^{m}=\int \limits _{-1}^{1}P_{k}^{m}\left(x\right)P_{l}^{m}\left(x\right)dx}
k
≠
l
.
{\displaystyle k\neq l.}
K
k
l
m
{\displaystyle K_{kl}^{m}}
k
=
l
:
{\displaystyle k=l:}
K
k
l
m
=
1
2
k
+
l
(
k
!
)
(
l
!
)
∫
−
1
1
{
(
1
−
x
2
)
m
d
k
+
m
d
x
k
+
m
[
(
x
2
−
1
)
k
]
}
{
d
l
+
m
d
x
l
+
m
[
(
x
2
−
1
)
l
]
}
d
x
.
{\displaystyle K_{kl}^{m}={\frac {1}{2^{k+l}\left(k!\right)\left(l!\right)}}\int \limits _{-1}^{1}\left\{\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right\}\left\{{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(x^{2}-1\right)^{l}\right]\right\}dx.}
Since
k
{\displaystyle k}
l
{\displaystyle l}
l
≥
k
.
{\displaystyle l\geq k.}
l
+
m
{\displaystyle l+m}
u
{\displaystyle u}
v
′
.
{\displaystyle v'.}
m
{\displaystyle m}
u
{\displaystyle u}
u
v
|
−
1
1
{\displaystyle \left.uv\right|_{-1}^{1}}
(
1
−
x
2
)
{\displaystyle \left(1-x^{2}\right)}
l
{\displaystyle l}
v
{\displaystyle v}
(
x
2
−
1
)
{\displaystyle \left(x^{2}-1\right)}
K
k
l
m
=
(
−
1
)
l
+
m
2
k
+
l
(
k
!
)
(
l
!
)
∫
−
1
1
(
x
2
−
1
)
l
d
l
+
m
d
x
l
+
m
[
(
1
−
x
2
)
m
d
k
+
m
d
x
k
+
m
[
(
x
2
−
1
)
k
]
]
d
x
.
{\displaystyle K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{k+l}\left(k!\right)\left(l!\right)}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]dx.}
Expand the second factor using Leibnitz' rule:
d
l
+
m
d
x
l
+
m
[
(
1
−
x
2
)
m
d
k
+
m
d
x
k
+
m
[
(
x
2
−
1
)
k
]
]
=
∑
r
=
0
l
+
m
(
l
+
m
)
!
r
!
(
l
+
m
−
r
)
!
d
r
d
x
r
[
(
1
−
x
2
)
m
]
d
l
+
k
+
2
m
−
r
d
x
l
+
k
+
2
m
−
r
[
(
x
2
−
1
)
k
]
.
{\displaystyle {\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]=\sum \limits _{r=0}^{l+m}{\frac {\left(l+m\right)!}{r!\left(l+m-r\right)!}}{\frac {d^{r}}{dx^{r}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{l+k+2m-r}}{dx^{l+k+2m-r}}}\left[\left(x^{2}-1\right)^{k}\right].}
The leftmost derivative in the sum is non-zero only when
r
≤
2
m
{\displaystyle r\leq 2m}
m
≤
l
{\displaystyle m\leq l}
k
+
l
+
2
m
−
r
≤
2
k
{\displaystyle k+l+2m-r\leq 2k}
r
≥
2
m
+
(
l
−
k
)
.
{\displaystyle r\geq 2m+(l-k).}
l
≥
k
{\displaystyle l\geq k}
r
=
2
m
{\displaystyle r=2m}
l
=
k
.
{\displaystyle l=k.}
K
k
l
m
=
(
−
1
)
l
+
m
2
2
l
(
l
!
)
2
(
l
+
m
)
!
(
2
m
)
!
(
l
−
m
)
!
δ
k
l
∫
−
1
1
(
x
2
−
1
)
l
d
2
m
d
x
2
m
[
(
1
−
x
2
)
m
]
d
2
l
d
x
2
l
[
(
1
−
x
2
)
l
]
d
x
.
{\displaystyle K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(l+m\right)!}{\left(2m\right)!\left(l-m\right)!}}\delta _{kl}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{2m}}{dx^{2m}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{2l}}{dx^{2l}}}\left[\left(1-x^{2}\right)^{l}\right]dx.}
To evaluate the differentiated factors, expand
(
1
−
x
2
)
k
{\displaystyle \left(1-x^{2}\right)^{k}}
(
1
−
x
2
)
k
=
∑
j
=
0
k
(
k
j
)
(
−
1
)
k
−
j
x
2
(
k
−
j
)
.
{\displaystyle \left(1-x^{2}\right)^{k}=\sum \limits _{j=0}^{k}\left({\begin{array}{c}k\\j\end{array}}\right)\left(-1\right)^{k-j}x^{2\left(k-j\right)}.}
2
k
{\displaystyle 2k}
x
2
k
{\displaystyle x^{2k}}
(
−
1
)
k
(
k
0
)
(
2
k
)
!
=
(
−
1
)
k
(
2
k
)
!
{\displaystyle \left(-1\right)^{k}\left({\begin{array}{c}k\\0\end{array}}\right)\left(2k\right)!=\left(-1\right)^{k}\left(2k\right)!}
K
k
l
m
=
1
2
2
l
(
l
!
)
2
(
2
l
)
!
(
l
+
m
)
!
(
l
−
m
)
!
δ
k
l
∫
−
1
1
(
x
2
−
1
)
l
d
x
{\displaystyle K_{kl}^{m}={\frac {1}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(2l\right)!\left(l+m\right)!}{\left(l-m\right)!}}\delta _{kl}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx}
Evaluate
∫
−
1
1
(
x
2
−
1
)
l
d
x
{\displaystyle \int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx}
x
=
cos
θ
⇒
d
x
=
−
sin
θ
d
θ
a
n
d
1
−
x
2
=
sin
θ
.
{\displaystyle x=\cos \theta \Rightarrow dx=-\sin \theta d\theta \;and\;1-x^{2}=\sin \theta .}
∫
−
1
1
(
x
2
−
1
)
l
d
x
=
∫
0
π
(
sin
θ
)
2
l
+
1
d
θ
.
{\displaystyle \int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta .}
π
→
0
{\displaystyle \pi \rightarrow 0\;}
0
→
π
{\displaystyle \;0\rightarrow \pi }
−
1
=
cos
(
π
)
{\displaystyle \;-1=\cos(\pi )\;}
1
=
cos
(
0
)
{\displaystyle \;1=\cos(0)\;}
A table of standard trigonometric integrals shows:
∫
0
π
sin
n
θ
d
θ
=
−
sin
θ
cos
θ
|
0
π
n
+
(
n
−
1
)
n
∫
0
π
sin
n
−
2
θ
d
θ
.
{\displaystyle \int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left.-\sin \theta \cos \theta \right|_{0}^{\pi }}{n}}+{\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta .}
−
sin
θ
cos
θ
|
0
π
=
0
,
{\displaystyle \left.-\sin \theta \cos \theta \right|_{0}^{\pi }=0,}
∫
0
π
sin
n
θ
d
θ
=
(
n
−
1
)
n
∫
0
π
sin
n
−
2
θ
d
θ
{\displaystyle \int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta }
n
≥
2.
{\displaystyle n\geq 2.}
∫
0
π
(
sin
θ
)
2
l
+
1
d
θ
{\displaystyle \int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta }
x
{\displaystyle x}
∫
−
1
1
(
x
2
−
1
)
l
d
x
=
2
(
l
+
1
)
2
l
+
1
∫
−
1
1
(
x
2
−
1
)
l
−
1
d
x
{\displaystyle \int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx}
l
≥
1.
{\displaystyle l\geq 1.}
∫
−
1
1
(
x
2
−
1
)
l
d
x
=
2
(
l
+
1
)
2
l
+
1
2
(
l
)
2
l
−
1
2
(
l
−
1
)
2
l
−
3
.
.
.
2
(
2
)
3
(
2
)
=
2
l
+
1
l
!
(
2
l
+
1
)
!
2
l
l
!
=
2
2
l
+
1
(
l
!
)
2
(
2
l
+
1
)
!
.
{\displaystyle \int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}{\frac {2\left(l\right)}{2l-1}}{\frac {2\left(l-1\right)}{2l-3}}...{\frac {2\left(2\right)}{3}}\left(2\right)={\frac {2^{l+1}l!}{\frac {\left(2l+1\right)!}{2^{l}l!}}}={\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}.}
Applying this result to (1):
K
k
l
m
=
1
2
2
l
(
l
!
)
2
(
2
l
)
!
(
l
+
m
)
!
(
l
−
m
)
!
2
2
l
+
1
(
l
!
)
2
(
2
l
+
1
)
!
δ
k
l
=
2
2
l
+
1
(
l
+
m
)
!
(
l
−
m
)
!
δ
k
l
.
{\displaystyle K_{kl}^{m}={\frac {1}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(2l\right)!\left(l+m\right)!}{\left(l-m\right)!}}{\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}\delta _{kl}={\frac {2}{2l+1}}{\frac {\left(l+m\right)!}{\left(l-m\right)!}}\delta _{kl}.}
![{\displaystyle P_{l}^{m}\left(x\right)={\frac {\left(-1\right)^{m}}{2^{l}l!}}\left(1-x^{2}\right)^{\frac {m}{2}}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(x^{2}-1\right)^{l}\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19c09e1bc89202d81456450622a6e522e26be8a1)
![{\displaystyle \left(\left[1-x^{2}\right]y^{'}\right)^{'}+\left(l\left[l+1\right]-{\frac {m^{2}}{1-x^{2}}}\right)y=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b93f67e55c4f1e910acf0a5ab798d40e4540c4eb)
![{\displaystyle K_{kl}^{m}={\frac {1}{2^{k+l}\left(k!\right)\left(l!\right)}}\int \limits _{-1}^{1}\left\{\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right\}\left\{{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(x^{2}-1\right)^{l}\right]\right\}dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dc0f6633dbc34f81b9194cb0a41598a95b44000)
![{\displaystyle K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{k+l}\left(k!\right)\left(l!\right)}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e595f931a7db501ce717e143822aa3468794f0ec)
![{\displaystyle {\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]=\sum \limits _{r=0}^{l+m}{\frac {\left(l+m\right)!}{r!\left(l+m-r\right)!}}{\frac {d^{r}}{dx^{r}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{l+k+2m-r}}{dx^{l+k+2m-r}}}\left[\left(x^{2}-1\right)^{k}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d9cbb431e752d17667fe7cde7bb777bd6eca15b)
![{\displaystyle K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(l+m\right)!}{\left(2m\right)!\left(l-m\right)!}}\delta _{kl}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{2m}}{dx^{2m}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{2l}}{dx^{2l}}}\left[\left(1-x^{2}\right)^{l}\right]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d80e2d7025fe8ac9ef821c5f991722f54809708)
