Why Does the Integral of Legendre Polynomials Yield a Kronecker Delta?

[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!

 

[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

 

[vanhees71]

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.