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I'm taking an ODE course at the moment, and my instructor gave us the following problem:

Derive the following formula for Legendre functions $Q_n(x)$ of the second kind:

$Q_n(x) = P_n(x) \int \frac{1}{[P_n(x)]^2 (1-x^2)}dx$

where $P_n(x)$ is the $n$-th Legendre polynomial.

He introduced Legendre functions in the context of second order ODEs, but we haven't really used them for anything - moreover, this is the only problem we were assigned that has anything to do with them. As a result, I'm sort of at a loss of where to start.

I've tried a couple of things (like using the actual Legendre ODE

$(1-x^2)y^{\prime \prime} - 2xy^{\prime} + n(n+1)y = 0$

and plugging in the solution $y(x)=a_1P_n(x)+a_2Q_n(x)$ and proceeding from there) but so far, haven't been able to go anywhere.

Any help (preferably as elementary as possible) would be much appreciated. Thanks!

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    Well, I've provided the question verbatim. I managed to find this proof of what I'm loo$k$ing for, but it's approximately five pages long and I'm almost entirely sure my instructor is looking for something simpler. For those interested, here is that proof: http://books.google.com/books?hl=en&id=Jj5pXGTZIKkC&pg=PA70. I'm combing through it for something I can use.2012-10-28

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The most general solution of Legendre equation is $y = A{P_n} + B{Q_n}.$ Let $y(x) = A(x){P_n}(x)$. Then $y' = AP' + A'P$ and $y'' = AP'' + 2A'P' + A''P$. So $(1 - {x^2})(AP'' + 2A'P' + A''P) - 2x(AP' + A'P) + n(n + 1)AP = 0.$ Note that $(1 - {x^2})(AP'') - 2x(AP') + n(n + 1)AP = 0$ which means some terms in the above equation vanish. Now let $A' = u$ and reduce the order so that $2\frac{{dP}}{P} + \frac{{du}}{u} - \frac{{2xdx}}{{1 - {x^2}}} = 0$ and $u = \frac{{{\text{const}}}}{{(1 - {x^2}){P^2}}}$ so $A = {C_n}\int {\frac{1}{{(1 - {x^2}){P^2}}}dx}.$ See if you can do the rest.

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    We could write $A = {C_n}\int_0^x {\frac{1}{{(1 - {y^2}){{\left[ {{P_n}(y)} \right]}^2}}}} dy$ and define ${C_0} = 1$ and ${D_0} = 0$ to avoid confusion.2012-10-28