# Legendre Polynomials!

I need some help to show that $\int_{-1}^{1}P_n^2(x)\ dx = \frac{2}{2n+1}$ where $P_n$ is the $n^{th}$ Legendre's Polynomial. I've already done some calculations, so I would be very grateful whether someone just showed me why:

$\int_0^1 s^n(1-s)^n ds = \frac{(n!)^2}{(2n+1)!}$

It would be enough :-)

P.S. The "reduction formula" can be used... (?)

• Do you have that $p(x) = P_n(x) - \frac{2n-1}{n}xP_{n-1}(x)$? Commented Sep 26, 2014 at 21:14
• So you have seen this and want to know why the Beta function values arise? Commented Sep 26, 2014 at 21:14
• an elementary way would be to substitute and apply this or that. Commented Sep 26, 2014 at 21:52
• AaronMaroja, I don't know if I have that, @ccorn Yes, I was looking at that but I have no need to change the variable, I guess.. Commented Sep 26, 2014 at 21:56

To start, note that the identity you are trying to prove is related to: $$\mathrm{B}(a, b) = \int_0^1 s^{a-1}(1-s)^{b-1}\,ds = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

(Your particular scenario is $\mathrm{B}(n, n)$.)

I'm sure there are many proofs of this, but one that I know relates to the convolution integral and Laplace transforms.

Declarations: Let $f(t) = t^{a-1},\,g(t) = t^{b-1}$. Also, let $h(t) = (f*g)(t)$, where $*$ is convolution. Finally, designate the Laplace transform of $f(t),\,g(t),\,h(t)$ as $F(s),\,G(s)\,H(s)$, respectively.

Note that $H(s) = F(s)G(s)$. (This is a property of convolution and the Laplace transform.)

By deriving or looking up the Laplace transforms of $f$ and $g$, we see that: \begin{align} H(s) &= F(s)G(s) \\ &= \frac{\Gamma(a)}{s^a}\frac{\Gamma(b)}{s^b} \\ &= \frac{\Gamma(a)\Gamma(b)}{s^{a+b}} \end{align}

Taking the inverse transform: \begin{align} h(t) &= \mathcal{L}^{-1}\left\{\frac{\Gamma(a)\Gamma(b)}{s^{a+b}}\right\}\\ &= \Gamma(a)\Gamma(b)\mathcal{L}^{-1}\left\{\frac{1}{s^{a+b}}\right\}\\ &= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left\{\frac{\Gamma(a+b)}{s^{a+b}}\right\}\\ &= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}t^{a+b-1} \end{align} By definition of convolution, we have that: $$h(t) = (f*g)(t) = \int_0^t r^{a-1}(t-r)^{b-1}\,dr$$

But, we know $h(t)$ from the inverse transform. This leads us to the identity: $$\int_0^t r^{a-1}(t-r)^{b-1}\,dr = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}t^{a+b-1}$$

For the Beta function, we substitute $t=1$, obtaining: $$\mathrm{B}(a,b) = \int_0^1 r^{a-1}(1-r)^{b-1}\,dr = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

Thus, for your case: \begin{align} \int_0^1 r^{n}(1-r)^{n}\,dr &= \frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma((n+1)+(n+1))}\\ &= \frac{n!n!}{(2n+1)!}\\ &= \frac{(n!)^2}{(2n+1)!} \end{align}

This is a "guided" exercise in Boyce and DiPrima's Ordinary Differential Equations and Boundary Value Problems, in the Convolution Integral section of the Laplace Transforms chapter. (I'm mentioning this to give credit to where I first saw this.)

• Thank you, but is there a way to prove that without using the "Beta function"? The paper I'm reading suggests using a "reduction formula", do you know what does that mean? Commented Sep 26, 2014 at 21:58
• I don't know to which reduction formula they are referring. I actually know nothing about Legendre polynomials, but just recognized the beta function. :) Commented Sep 26, 2014 at 22:05

Instead of calculating $I_n=\int_0^1s^n(1-s)^nds$ I propose to calculate $J_{n,m}=\int_0^1s^n(1-s)^mds$ and then we set $m=n$. With this said we have using integration by parts, for $m>0$: $$J_{n,m}=\left[\frac{s^{n+1}}{n+1}(1-s)^{m}\right]_0^1+\int_0^1s^{n+1}(1-s)^{m-1}ds$$ or $$J_{n,m}=\frac{m}{n+1}J_{n+1,m-1}$$ This shows by induction that $$J_{n,m}=\frac{m}{n+1}\cdot \frac{m-1}{n+2}\cdots \cdot \frac{1}{n+m}J_{n+m,0} =\frac{m!\,n!}{(m+n+1)!}$$ Now, set $m=n$.

• +1 Much, much nicer than my way. :) This is probably what the book was looking for, too, given that the recurrence for the value of the integral is sometimes called a reduction formula. Commented Sep 27, 2014 at 17:23
• @anorton, Yes, the formula for $B(n,m)=J(n+1,m+1)$ in terms of factorials is readily proved by induction for positive integer arguments. Commented Sep 27, 2014 at 17:36