How do I find the normalizing factor for the Legendre polynomials from the generating function?
The generating function is:
$$
\frac{1}{\sqrt{1-2xh+h^2}}= \sum^{\infty}_{l=0}P_lh^l.
$$
I am doing this from Boas, 12.23.1. There is a hint that I have tried to pursue: square this equation and integrate from -1,1 then expand the integral in powers of h and equate the coefficients.
My Work so Far
Squaring both sides and integrating $x$ (assuming that I should do this to x) from $-1,\text{ to } 1$.
$$
\int^1_{-1} \frac{dx}{1-2xh+h^2}=\int^1_{-1}dx(P_0^2 + hP_0P_1 +…+h^2P_1^2+h^3P_1P_2+…+…).
$$
Because all $\int^1_{-1}P_n(x)P_m(x)dx=0$, when $n\neq m$ we can get rid of a lot of terms.
$$
\int^1_{-1} \frac{dx}{1-2xh+h^2}=\int^1_{-1}\sum^{\infty}_{l=0}h^{2l}P_l(x)^2 dx.
$$
Any help will be greatly appreciated. I am stumped on integrating.
Attempt at Integrating the left hand side.
Set $u=1-2xh+h^2\Rightarrow du=-2hdx$:
\begin{align}
-&2h\int^{1-2h+h^2}_{1+2h+h^2} \frac{du}{u}\\
= -&2h\ln(u)|^{1-2h+h^2}_{1+2h+h^2}\\
=-&2h\ln\left(\frac{1-2h+h^2}{1+2h+h^2}\right)
\end{align}
… But this does not feel correct.
$$
-2h\ln\left(\frac{1-2h+h^2}{1+2h+h^2}\right)=\int^1_{-1}\sum^{\infty}_{l=0}h^{2l}P_l(x)^2 dx
$$
With a little help from my friends
$$
4h\ln\left(\frac{1+h}{1-h}\right)=\left[\ln(1+h)-\ln(1-h) \right]\\
=4h\left[(h-\frac{h^2}{2}+\frac{h^3}{3}-\frac{h^4}{4}+…-…)-(-h -\frac{h^2}{2}-\frac{h^3}{3}-\frac{h^4}{4}-…) \right]\\
=8h\left[h+\frac{h^3}{3}+\frac{h^5}{5}… \right]\\
=8h\sum^\infty_{n=0}\frac{h^{2n+1}}{2n+1} = \int^1_{-1}dx\sum^{\infty}_{l=0}h^{2l}P_l(x)^2 dx.
$$
I think I divide both sides by $\sum^{\infty}_{l=0}h^{2l}$:
$$
\Rightarrow 8h\sum^\infty_{n=0}\frac{h}{2n+1}= 8h^2\sum^\infty_{n=0}(2n+1)^{-1}
$$
But this sum diverges…
Best Answer
You made a small arithmetic error during your substitution, which when corrected should give you the correct result. Using the same substitution $u = 1 - 2xh + h^2 \implies \frac{\mathrm{d}u}{-2h} = \mathrm{d}x$
\begin{align*} \int_{-1}^1 \frac{1}{1-2xh + h^2} \mathrm{d}x &= \frac{1}{-2h} \int_{(1+h)^2}^{(1-h)^2} \frac{\mathrm{d}u}{u} \\ &= -\frac{1}{2h}\log\frac{(1-h)^2}{(1+h)^2} \\ &= \frac{1}{h} (\log(1+h) - \log(1-h)) \\ &= \frac{1}{h} \left(\sum_{n=1}^\infty \frac{h^n}{n} - \frac{(-h)^n}{n}\right) \\ &= \frac{1}{h} \sum_{n\ \mathrm{odd}} \frac{2h^n}{n} \\ &= \frac{2}{h} \sum_{l = 0}^\infty \frac{h^{2l+1}}{2l+1} \\ &= \sum_{l = 0}^\infty \frac{2h^{2l}}{2l+1} \end{align*} Equating this to the right hand side, we have
$$\sum_{l = 0}^\infty \frac{2h^{2l}}{2l+1} = \sum_{l=0}^\infty h^{2l}\int_{-1}^1 P_l^2 \mathrm{d}x$$ Equating coefficients gives the desired result: $$ \int_{-1}^1 P_n^2 \mathrm{d}x = \frac{2}{2n+1}$$