3

In addition to a former question. I wish to show that:

$$\int_{-1}^1 x^n P_n (x) dx = \frac{n}{2n+1}\int_{-1}^1 x^{n-1} P_{n-1} (x) dx$$

Assuming the result of:

$$P'_{n+1}-P'_{n-1} = (2n+1)P_{n}$$

Multiplying both sides with $x^n$. We obtain:

$$x^nP'_{n+1}-x^nP'_{n-1} = x^n(2n+1)P_{n}$$

Now integrating both sides:

$$\int_{-1}^1x^nP'_{n+1}-x^nP'_{n-1}dx = \int_{-1}^1 x^n(2n+1)P_{n}dx$$

Now have to integrate by parts on the lefthand side. This becomes:

$$\int_{-1}^1 x^n(P'_{n+1}-P'_{n-1})dx= x^n[P_{n+1}-P_{n-1}]^{1}_{-1} - \int_{-1}^1 nx^{n-1}(P_{n+1}-P_{n-1})dx$$But here I am stuck. Any further help appreciated again.

mathpieuler
  • 325
  • Multiply your formula by $x^n$ and integrate over $[-1,1]$ using integration by parts on the left hand side. For the first integral note that $P_n$ is an even/odd function when $n$ is even/odd. – Winther Dec 26 '17 at 15:09
  • I tried to use your hint, and added it to the original question. But I am stuck halfway.... – mathpieuler Dec 26 '17 at 21:43
  • The first term $[\cdots]{-1}^1$ is zero (since $P_n(1) = 1$ and $P_n(-1) = (-1)^n$) and using the relation $P{n+1} = \frac{(2n+1)}{n+1}xP_n - \frac{n}{n+1}P_{n-1}$ on the $x^{n-1}P_{n+1}$ term gives the result. – Winther Dec 28 '17 at 00:26

1 Answers1

1

Hint. By using Rodrigues' formula, $$ P_n(x)=\frac{1}{n!\space2^n}\frac{d^n}{dx^n}(x^2-1)^n, $$ one may integrate by parts $n$ times obtaining $$ \begin{align} \int_{-1}^{1} x^{n}P_{n}(x) dx&=\frac{(-1)^n}{n!\space2^n}\cdot n!\int_{-1}^{1} (x^2-1)^n dx \\\\&=\frac{1}{2^n}\int_0^{1} u^{-1/2}(1-u)^n du \end{align} $$ that is, using the Euler beta function we have

$$ \int_{-1}^{1} x^{n}P_{n}(x) \:dx=\frac{2^n\:(n!)^2}{(2n+1)!}. $$

Olivier Oloa
  • 120,989
  • So I don't have to use $P'{n+1}-P'{n-1} = (2n+1)P_{n}$. Because the above result can be deduced by my statement which I wish to prove ie $\int_{-1}^1 x^n P_n (x) dx = \frac{n}{2n+1}\int_{-1}^1 x^{n-1} P_{n-1} (x) dx$ – mathpieuler Dec 26 '17 at 15:14