Legrendre integral using Rodriquez' formula

1. Oct 9, 2015

ognik

1. The problem statement, all variables and given/known data
Show that $$\int_{-1}^{1}{x}^{n}P_n(x) \,dx= \frac{2^{n+1}n!n!}{(2n+1)}!$$
(Side question: this is the orthogonality integral= 0 when the functions inside the integrand are orthogonal; but I have seen hints that sometimes the Integral is NOT=0 even though the functions are orthogonal - would appreciate confirmation and understanding of that please?)
2. Relevant equations
Hint: Use Rodriquez formula $$P_n(x)=\frac{1}{2n}\frac{1}{n!} \frac{d ^{n}({x}^{2}-1)^n} {dx^{n}}$$
3. The attempt at a solution
(NB: this section is new to me, my aim is to do & fully understand every problem in the book, but I have deadlines -so need some help where I get a bit bogged down. Please be vigilant as to where I might have understood some theory wrong or missed something)
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To simplify the workings, the limits are [-1,1] throughout, I kept aside this term in n for now: $$\frac{1}{2n}\frac{1}{n!}; \:and\:also\:let\: (x^2-1) = f$$
Then, by parts, $$\int {x}^{n} \frac{ d{^{n}{f}^n}} {{dx}^{n}} dx = {x}^{n} \frac{ d{^{n-1}{f}^n}} {{dx}^{n-1}} |^{1}_{-1} - \int n{x}^{n-1} \frac{ d{^{n-1}{f}^n}} {{dx}^{n-1}} dx$$ - and repeating by parts n times. I then argue as follows (and would appreciate improvements/corrections as appropriate)

1st term (outside integral): After doing 3 Successive differentiations of $$\frac{ d{^{n-1}{f}^n}} {{dx}^{n-1}}, - \: ALL \:have \:a \: ({x}^{2}-1) \: term \:in \:them =(x+1)(x-1)$$
Therefore I claim all the following terms over [-1,1] will vanish, ie $$\frac{d{^{i}{f}^n}}{{x}^{i}} =0$$

I also claim that further integration by parts will always produce a 1st term (outside the integral) which will include $$\frac{d{^{i}{f}^n}}{{x}^{i}}$$, and to which the above claim applies. So, comfortable that the 1st(outer terms) will always vanish through repeated integration by parts, the problem is simplified to the sequence of the integral part only.
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The Integrand will always have 2 parts, A) and B):
$$A) x^n$$ This part is repeatedly differentiated with each integration (always making it the 'u' term for parts). Therefore after n integrations it becomes $$n!x^{0}=n!$$

$$B) \frac {d{^{n}{f^n}}} {{dx}^{n}}$$ Because this is always 'dv' by parts, each successive integral just produces a reduced derivative, until finally we are left with just $$f^n$$
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So putting it all together, I can find $$\frac{1}{2n}\frac{1}{n!} n! \int_{-1}^{1}f^n \,dx = \frac{1}{2n} \frac{{(x^2-1)}^{n}}{n(n+1)}|^1_{-1}$$
Tantalisingly close, but after checking a few times, I still can't see what I have done wrong above?

2. Oct 10, 2015

MisterX

Your anti-derivative does not seem correct to me. For example $$\int f^1dx = \int (x^2 - 1)dx = \frac{1}{3}x^3 - x + C$$ Unless I am missing something, anti-derivative of $f^n$ cannot be written as $(x^2 -1)^n*g(n)$ the way you seem to have done.

3. Oct 11, 2015

ognik

Probably not, I think I used parts, something like splitting the integrand into $$(x^2-1)^{n-1}(x^2-1)$$

Anyway, I have subsequently stumbled upon the 'beta function' with which I made more progress, rewriting my eqtn as $$\frac{1}{2n} [2\int_0^1(1-x^2)^n dx$$

Then the part inside the [] should = $$B(\frac{1}{2}, n+1)$$
But, not having encountered Bet fanction before, I am struggling to turn this Beta function into $$\frac {2^{n+1}n!n!}{(2n+1)!}$$

(PS: Is there a way to write latex inline? ie not using )