# Legendre Polynomials question

Don'tKnowMuch
I am doing a Laplace's equation in spherical coordinates and have come to a part of the problem that has the integral...

∫ P(sub L)*(x) * P(sub L')*(x) dx (-1<x<1)

The answer to this integral is given by a Kronecker delta function (δ)...

= 0 if L ≠ L'

OR....
= 2/(2L+1)*δ if L = L' (where δ = 1)

I believe the reason why the integral is equal to zero when L ≠ L' is because of the orthogonality of Legendre polynomials, however i cannot figure out how the integral is equal 2/(2L+1). If L = L' then the integral would equal...

∫ [P(sub L)*(x)]^2 dx

I tried to use a simple power rule of integration to solve the above integral, but i am afraid my method is flawed. Any suggestions?

Pre-emptive thanks to whomever takes the time to help!!

## Answers and Replies

Don'tKnowMuch
For clarity i should add that...

P(sub L)(x)

is a Legendre polynomial

Don'tKnowMuch
I think i have to use Rodrigues' formula

Gold Member
2021 Award
This is one possibility. Another is to use the generating function, which is used to define in the msot convenient way the Legendre polynomials.

It's given by
$$\Phi(r,u)=\frac{1}{\sqrt{1-2u r+r^2}}=\sum_{l=0}^{\infty} P_l(u) r^l.$$
From this you find
$$P_l(u)=\frac{1}{l!} \left .\frac{\partial^l}{\partial r^l} \Phi(r,u) \right|_{r=0}.$$
This series expansion is obviously valid for $|r|<1$ if $|u|<1$.

We know that the Legendre Polynomials are orthogonal to each other since they are solutions of the eigen-value equation
$$\frac{\mathrm{d}}{\mathrm{d} u} \left [(1-u^2) \frac{\mathrm{d} P_l}{\mathrm{d} u} \right ]=-l(l+1) P_l,$$
where the differential operator is self-adjoint on $L^2([-1,1])$. Thus we have
$$\int_{-1}^{1} \mathrm{d} u P_l(u) P_{l'}(u)=N_l \delta_{ll'}.$$
To evaluate the normalization factor we take the following integral:
$$I(r)=\int_{-1}^{1} \mathrm{d} u \Phi^2(r,u)=\int_{-1}^{1} \mathrm{d} u \frac{1}{1-2 u r +r^2}=-\left [\frac{1}{2r} \ln(1-2 u r+r^2) \right]_{u=-1}^{u=1}=\frac{1}{r} \ln \left (\frac{1+r}{1-r} \right ).$$
Expanding this into a power series in $r$ yields
$$I(r)=\sum_{l=0}^{\infty} \frac{2}{2l+1} r^{2l}.$$
On the other hand from the definition of the Legendre Polynomials by the above given series expansion you get
$$\Phi^2(r,u)=\sum_{l,l'=0}^{\infty} P_l(u) P_{l'}(u) r^{l+l'}.$$
Integrating over $u$ leads to
$$I(r)=\sum_{l,l'=0}^{\infty} N_l \delta_{ll'} r^{2l}=\sum_{l=0}^{\infty} N_l r^{2l}.$$
Comparing the coefficients in both expansions of $I(r)$ yields
$$N_l=\frac{2}{2l+1}.$$
QED.