0
$\begingroup$

prove:

$\int_{-1}^1 (x^2P_{n+1}(x)+P_{n-1}(x))\ dx$=$2n(n+1)\over {(2n-1)(2n+1)(2n+3)}$

I think to use the formula:

$ (n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$

Then multiply the L.H.S. and R.H.S. by $x^2P_{n-1}(x)$ and integrate the two sides, so we get:

$\int_{-1}^1 (n+1)x^2P_{n+1}(x)\ dx=\int_{-1}^1 (2n+1)x^3P_n(x)P_{n-1}(x) \ dx-\int_{-1}^1 nx^2P^2_{n-1}(x) \ dx$

True? And if it is true how can I find the R.H.S. ?

$\endgroup$
  • $\begingroup$ Prove what? I don't see any statement to prove, just an integral. $\endgroup$ – Robert Israel Apr 5 '16 at 22:14
  • $\begingroup$ sorry, I edited this, Thanks. $\endgroup$ – Dima Apr 5 '16 at 22:16
  • $\begingroup$ Are you familiar with the technique known as Integration By Parts? And also, do you know what $\int P_{n}(x) dx$ equals indefinitely? $\endgroup$ – Xoque55 Apr 5 '16 at 22:27
  • $\begingroup$ You should study the orthogonality relation of Legendre polynomials. Anyway, your statement is wrong. Maybe some square is missing? $\endgroup$ – Jack D'Aurizio Apr 5 '16 at 23:47
0
$\begingroup$

There seems to be a typo in the original question. I am assuming it is $$\int_{-1}^1x^2P_{n+1}(x)\cdot P_{n-1}(x)dx=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$ With that correction, my first try would have been to apply the Rodrigues formula and integrate by parts $n+1$ times, and while that works, it seems easier to apply the recurrence relation for the Legendre polynomials, $$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$ To get $$xP_{n+1}(x)=\frac{(n+2)}{(2n+3)}P_{n+2}(x)+\frac{(n+1)}{(2n+3)}P_n(x)$$ and $$xP_{n-1}(x)=\frac{n}{(2n-1)}P_n(x)+\frac{(n-1)}{(2n-1)}P_{n-1}(x)$$ And then apply the orthogonality relation for the Legendre polynomials $$\int_{-1}^1P_n(x)P_m(x)dx=\frac2{(2n+1)}\delta_{nm}$$ So now only the $P_n(x)$ terms survive orthogonality and we have $$\int_{-1}^1x^2P_{n+1}(x)\cdot P_{n-1}(x)dx=\frac{(n+1)}{(2n+3)}\cdot\frac{n}{(2n-1)}\cdot\frac2{(2n+1)}=\frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$

$\endgroup$
0
$\begingroup$

It is enough to exploit the following definition of Legendre polynomials:

The sequence $\{P_n(x)\}_{n\geq 0}$ has the property that $\partial P_n(x)=n$ and $$ \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2\cdot \delta(n,m)}{2n+1}.$$

Since $P_0(x)=1, P_1(x)=x$ and $P_2(x)=\frac{3}{2}x^2-\frac{1}{2}$, assuming $n\geq 1$:

$$\begin{eqnarray*} \int_{-1}^{1}\left(x^2 P_{n+1}(x)+P_{n-1}(x)\right)\,dx &=&2\cdot\delta(0,n-1)+\int_{-1}^{1}\frac{2P_2(x)+P_0(x)}{3}\cdot P_{n+1}(x)\,dx \\&=&2\cdot \delta(n,1)+\frac{4}{15}\cdot\delta(2,n+1)+\frac{2}{3}\cdot\delta(0,n+1)\end{eqnarray*}$$ hence your integral equals $2$ is $n=0$, $\frac{34}{15}$ if $n=1$ and $0$ otherwise.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.