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ThePerfectHacker · #6 July 29th, 2007, 03:30 PM

As mentioned this problem can be solved with Legendre Polynomials and their identities, but there is no need. There are many nice identities among the Legendre Polynomials and not a single one is used here.

The trick is to write the differencial equation as (in Sturm-Louiville form),
$[(1-x^2)y']'+n(n+1)y=0$

Thus,
$[(1-x^2)P_n'(x)]'+n(n+1)P_n(x)=0$
Multiply by $P_m(x)$ to get,
$[(1-x^2)P_n'(x)]'P_m(x)+n(n+1)P_n(x)P_m(x)=0$
And integrate,
$\int_{-1}^1 [(1-x^2)P_n'(x)]'P_m(x) dx + n(n+1)\int_{-1}^1 P_n(x)P_m(x)dx=0$
Use integration by part on the first integral,
$(1-x^2)P_n'(x)P_m(x)\big|_{-1}^1 - \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
The first term vanishes,
$- \int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+n(n+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
Doing the similar steps with $P_m(x)$ instead we have,
$-\int_{-1}^1(1-x^2)P_n'(x)P_m'(x)dx+m(m+1)\int_{-1}^1P_n(x)P_m(x)dx=0$
Now subtract these two equations to get,
$[n(n+1)-m(m+1)]\int_{-1}^1 P_n(x)P_m(x) dx = 0$
Since $n\not = m$ and they are integers we have,
$\int_{-1}P_n(x)P_m(x) dx = 0$ .
Q.E.D.