F. Norm of the Legendre polynomials

To compute the norm of the solution of the equation for the geometry, we need to compute the norm of the associated Legendre polynomial, which is defined as follows

∫−11[Pk(m)(z)]2 dz,\int_{-1}^1 \left[ P_k^{(m)}(z) \right]^2 \,dz,
(F.1)

where Pk(m)(z)P_k^{(m)}(z) =(1−z2)m2⋅dmdzmPk(z)= (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) is the associated Legendre polynomial. Let us substitute this definition into the integral (F.1):

∫−11[Pk(m)(z)]2 dz\displaystyle \int_{-1}^1 \left[ P_k^{(m)}(z) \right]^2 \,dz =∫−11(1−z2)m[dmdzmPk(z)]2 dz.\displaystyle = \int_{-1}^1 (1 - z^2)^m \left[ \frac{d^m}{dz^m} P_k(z) \right]^2 \,dz.

Let us substitute the Rodrigues formula (E.2) into the integral and take out everything that does not depend on zz:

∫−11(1−z2)m[dmdzmPk(z)]2 dz\displaystyle \int_{-1}^1 (1 - z^2)^m \left[ \frac{d^m}{dz^m} P_k(z) \right]^2 \,dz =122k⋅(k!)2⋅∫−11(1−z2)m⋅[dk+mdzk+m(z2−1)k]2 dz.\displaystyle = \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 (1 - z^2)^m \cdot \left[ \frac{d^{k+m}}{dz^{k+m}} \left(z^2 - 1 \right)^k \right]^2 \,dz.
(F.2)

Clearly, mm ≤k\leq k: otherwise the order of the derivative kk +m+ m exceeds the degree of the polynomial 2⋅k2 \cdot k, and the integrand is identically zero. First let us compute the auxiliary integral corresponding to the case mm =0= 0

122k⋅(k!)2⋅∫−11[dkdzk(z2−1)k]2 dz.\frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz.
(F.3)

Let us integrate it by parts:

∫−11[dkdzk(z2−1)k]2 dz\displaystyle \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz =∣u=dkdzk(z2−1)k,du=dk+1dzk+1(z2−1)k dzdv=dkdzk(z2−1)k dz,v=dk−1dzk−1(z2−1)k∣\displaystyle = \left | \begin{array}{ll} u = \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k, &du = \frac{d^{k+1}}{dz^{k+1}} \left(z^2 - 1 \right)^k \,dz\\ dv = \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \,dz, &v = \frac{d^{k-1}}{dz^{k-1}} \left(z^2 - 1 \right)^k \end{array} \right |

The boundary terms are zero: the derivatives didzi(z2−1)k\frac{\displaystyle d^i}{\displaystyle dz^i} \left(z^2 - 1 \right)^k for ii <k< k vanish at the endpoints of the interval. After applying the integration by parts formula kk times (each application changes the sign) we obtain

122k⋅(k!)2⋅∫−11[dkdzk(z2−1)k]2 dz\displaystyle \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz =(−1)k22k⋅(k!)2⋅∫−11(z2−1)k⋅d2kdz2k(z2−1)k dz,\displaystyle = \frac{\displaystyle (-1)^k}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left(z^2 - 1 \right)^k \cdot \frac{d^{2k}}{dz^{2k}} \left(z^2 - 1 \right)^k \,dz,

now from Newton's binomial formula it becomes obvious that d2kdz2k(z2−1)k\frac{d^{2k}}{dz^{2k}} \left(z^2 - 1 \right)^k =(2k)!= (2k)!: the polynomial has maximal degree 2⋅k2 \cdot k, a derivative of that order kills all terms except the leading one, and its coefficient equals (2k)!(2k)!.

It remains to compute the integral

∫−11(z2−1)k dz\displaystyle \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =∣u=(z2−1)k,du=2⋅k⋅z⋅(z2−1)k−1 dzdv=dz,v=z∣\displaystyle = \left | \begin{array}{ll} u = \left(z^2 - 1 \right)^k, &du = 2 \cdot k \cdot z \cdot (z^2 - 1)^{k-1} \,dz\\ dv = dz, &v = z \end{array} \right |

obviously, the boundary term equals zero, therefore

∫−11(z2−1)k dz\displaystyle \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =−2⋅k⋅∫−11z2⋅(z2−1)k−1 dz,\displaystyle = -2 \cdot k \cdot \int_{-1}^1 z^2 \cdot (z^2 - 1)^{k-1} \,dz,

let us replace z2z^2 by z2z^2 −1- 1 +1+ 1 and obtain

∫−11(z2−1)k dz\displaystyle \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =−2⋅k2⋅k+1⋅∫−11(z2−1)k−1 dz,\displaystyle = - \frac{\displaystyle 2 \cdot k}{\displaystyle 2 \cdot k + 1} \cdot \int_{-1}^1 \left(z^2 - 1 \right)^{k-1} \,dz,

applying this recurrence formula another kk −1- 1 times, we obtain

∫−11(z2−1)k dz\displaystyle \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =(−1)k⋅2⋅k2⋅k+1⋅2⋅(k−1)2⋅(k−1)+1⋅2⋅(k−2)2⋅(k−2)+1⋅⋯⋅2⋅12⋅1+1⋅2,\displaystyle = (-1)^k \cdot \frac{\displaystyle 2 \cdot k}{\displaystyle 2 \cdot k + 1} \cdot \frac{\displaystyle 2 \cdot (k - 1)}{\displaystyle 2 \cdot (k - 1) + 1} \cdot \frac{\displaystyle 2 \cdot (k - 2)}{\displaystyle 2 \cdot (k - 2) + 1} \cdot \cdots \cdot \frac{\displaystyle 2 \cdot 1}{\displaystyle 2 \cdot 1 + 1} \cdot 2,
∫−11(z2−1)k dz\displaystyle \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =(−1)k⋅2⋅2k⋅k!⋅2k⋅k!(2⋅k+1)!.\displaystyle = (-1)^k \cdot 2 \cdot \frac{\displaystyle 2^k \cdot k! \cdot 2^k \cdot k!}{\displaystyle (2 \cdot k + 1)!}.

As a result, the factors (−1)k(-1)^k cancel, and we obtain

122k⋅(k!)2⋅∫−11[dkdzk(z2−1)k]2 dz\displaystyle \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz =22⋅k+1.\displaystyle = \frac{\displaystyle 2}{\displaystyle 2 \cdot k + 1}.
(F.4)

Let us write the well-known recurrence formula for the associated Legendre polynomials, which will be needed for the next integration by parts

2⋅m⋅z⋅Pk(m)(z)\displaystyle 2 \cdot m \cdot z \cdot P_k^{(m)}(z) =1−z2⋅[Pk(m+1)(z)+(k+m)⋅(k−m+1)⋅Pk(m−1)(z)],\displaystyle = \sqrt{1 - z^2} \cdot \left[ P_k^{(m+1)}(z) + (k+m) \cdot (k-m+1) \cdot P_k^{(m-1)}(z) \right],
(F.5)

for a slight simplification let us denote ϵ\epsilon =(k+m)⋅(k−m+1)= \left( k + m \right) \cdot \left( k - m + 1 \right).

Let us return to the integral (F.2) and perform integration by parts:

∫−11(1−z2)m⋅[dmdzmPk(z)]2 dz\displaystyle \int_{-1}^1 \left( 1 - z^2 \right)^m \cdot \left[ \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) \right]^2 \,dz =∣u=(1−z2)m⋅dmdzmPk(z)du=(1−z2)(m/2)−1⋅[−2⋅m⋅z⋅Pk(m)(z)+1−z2⋅Pk(m+1)(z)] dz=(1−z2)m−12⋅[−Pk(m+1)(z)−ϵ⋅Pk(m−1)(z)+Pk(m+1)(z)] dz=−(1−z2)m−12⋅ϵ⋅Pk(m−1)(z) dzdv=dmdzmPk(z) dz,v=dm−1dzm−1Pk(z)=(1−z2)1−m2⋅Pk(m−1)(z)∣\displaystyle = \left | \begin{aligned} &u = \left( 1 - z^2 \right)^m \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z)\\ &du = \left( 1 - z^2 \right)^{(m/2)-1} \cdot \left[ -2 \cdot m \cdot z \cdot P_k^{(m)}(z) + \sqrt{1 - z^2} \cdot P_k^{(m+1)}(z) \right] \,dz=\\ &\left( 1 - z^2 \right)^{\frac{\displaystyle m-1}{\displaystyle 2}} \cdot \left[ - P_k^{(m+1)}(z) - \epsilon \cdot P_k^{(m-1)}(z) + P_k^{(m+1)}(z) \right] \,dz=\\ &- \left( 1 - z^2 \right)^{\frac{\displaystyle m-1}{\displaystyle 2}} \cdot \epsilon \cdot P_k^{(m-1)}(z) \,dz\\ &dv = \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) \,dz,\\ &v = \frac{\displaystyle d^{m-1}}{\displaystyle dz^{m-1}} P_k(z) = \left( 1 - z^2 \right)^{\frac{\displaystyle 1-m}{\displaystyle 2}} \cdot P_k^{(m-1)}(z) \end{aligned} \right |

the boundary term contains the factor 1−z2\sqrt{1 - z^2} and vanishes at the endpoints of the interval, so we obtain

∫−11Pk(m)(z)⋅Pk(m)(z) dz\displaystyle \int_{-1}^1 P_k^{(m)}(z) \cdot P_k^{(m)}(z) \,dz =(k+m)⋅(k−m+1)⋅∫−11Pk(m−1)(z)⋅Pk(m−1)(z) dz,\displaystyle = (k+m) \cdot (k-m+1) \cdot \int_{-1}^1 P_k^{(m-1)}(z) \cdot P_k^{(m-1)}(z) \,dz,

successively applying integration by parts another mm −1- 1 times and taking into account (F.4), we obtain

∫−11Pk(m)(z)⋅Pk(m)(z) dz\displaystyle \int_{-1}^1 P_k^{(m)}(z) \cdot P_k^{(m)}(z) \,dz =22⋅k+1⋅(k+m)⋅(k−m+1)⋅(k+m−1)⋅(k−m+2)⋯ ,\displaystyle = \frac{\displaystyle 2}{\displaystyle 2 \cdot k + 1} \cdot (k+m) \cdot (k-m+1) \cdot (k+m-1) \cdot (k-m+2) \cdots,

the factors with a plus decrease from kk +m+ m to kk +1+ 1, the factors with a minus increase from kk −m- m +1+ 1 to kk (it is important here that mm ≤k\leq k holds), so

(k+m)⋅(k−m+1)⋅(k+m−1)⋅(k−m+2)⋯\displaystyle (k+m) \cdot (k-m+1) \cdot (k+m-1) \cdot (k-m+2) \cdots =(k+m)!k!⋅k!(k−m)!\displaystyle = \frac{\displaystyle (k+m)!}{\displaystyle k!} \cdot \frac{\displaystyle k!}{\displaystyle (k-m)!} =(k+m)!(k−m)!.\displaystyle = \frac{\displaystyle (k+m)!}{\displaystyle (k-m)!}.

The final norm of the associated Legendre polynomial has the form

∫−11[Pk(m)(z)]2 dz\displaystyle \int_{-1}^1 \left[ P_k^{(m)}(z) \right]^2 \,dz =22⋅k+1⋅(k+m)!(k−m)!,m\displaystyle = \frac{\displaystyle 2}{\displaystyle 2 \cdot k + 1} \cdot \frac{\displaystyle (k+m)!}{\displaystyle (k-m)!}, \quad m ≤k.\displaystyle \leq k.
(F.6)