# Relationship between Legendre polynomials and Legendre functions of the second kind

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!

• That formula makes no sense -- you're multiplying a function of $x$ by an indefinite integral over the bound variable $x$. Even if that's interpreted as again yielding a function of $x$, there's still an unspecified integration constant. You'll need to provide limits for the claim to be well-defined. – joriki Oct 28 '12 at 2:54
• Well, I've provided the question verbatim. I managed to find this proof of what I'm looking 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: books.google.com/books?hl=en&id=Jj5pXGTZIKkC&pg=PA70. I'm combing through it for something I can use. – Bachmaninoff Oct 28 '12 at 3:06

## 1 Answer

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.

• Thank you for the response! I'll certainly take a look at this. This looks substantially cleaner than another approach I saw in a textbook I've linked to in the comments. – Bachmaninoff Oct 28 '12 at 3:11
• By the way I think your formula is not general enough. We do choose some constants to be 0 or 1 for convenience, but I think the most general formula is: $${Q_n}(x) = {C_n}{P_n}(x)\int {\frac{1}{{(1 - {x^2}){{\left[ {{P_n}(x)} \right]}^2}}}dx + {D_n}{P_n}(x)}$$ which you can deduce from my answer. – glebovg Oct 28 '12 at 3:18
• @glebovg: You're carrying forward the confusion sown by the indefinite integral in the question. You wrote an indefinite integral and then said it's not the most general result, you can add a constant $D_n$ -- but the indefinite integral is only defined up to that constant anyway, so the constant was already there before you made it explicit. – joriki Oct 28 '12 at 3:33
• 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. – glebovg Oct 28 '12 at 3:46