1
$\begingroup$

I need the results of $\int_{-1}^1 p_i(x) p_j'(x) dx$, where $p_i$ is the classical Legendre polynomials in $[-1,1]$. Using the matlab, I get the following result:

i \ j 0 1 2 3 4
0 0 2 0 2 0
1 0 0 2 0 2
2 0 0 0 2 0
3 0 0 0 0 2
4 0 0 0 0 0

It seems that $\int_{-1}^1 p_{n}(x) p_{n+1+2k}'(x) dx =2 ,\quad n\geq 0,\ k \geq 0$. I am not familiar with Legendre polynomials. If this conclusion is correct, I want to know if there are some theorems to help me get this conclusion easily.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

This is only a partial answer and explains all the zeroes in the table, and the first diagonal of $2$s. I will try to update it soon to cover all the cases.

Recall the Legendre polynomials are $$P_n(x)=\frac1{2^nn!}\frac{\mathrm d^n}{\mathrm dx^n}(x^2-1)^n$$ Therefore, for any smooth function $f(x)$, $$\int_{-1}^1P_n(x)f(x)\,\mathrm dx$$ can be evaluated by repeatedly using integration by parts. $$\int_{-1}^1P_n(x)f(x)\,\mathrm dx=\left[\frac{1}{2^nn!}\frac{\mathrm d^{n-1}}{\mathrm dx^{n-1}}(x^2-1)^nf(x)\right]_{-1}^1-\int_{-1}^1\frac{1}{2^nn!}\frac{\mathrm d^{n-1}}{\mathrm dx^{n-1}}(x^2-1)^nf^{(1)}(x)\,\mathrm dx$$ The first term vanishes and the second term can be again integrated by parts. In the end, you will get $$\int_{-1}^1P_n(x)f(x)\,\mathrm dx=\frac{1}{2^nn!}\int_{-1}^1f^{(n)}(x)(1-x^2)^n\,\mathrm dx$$ Now, if $f(x)=P_m^\prime(x)$ for $m\leq n$, then $P_m^{(n+1)}=0$ so the integral vanishes. Also, $P_m^{(n+1)}$ is an odd function of $x$ if $m-n$ is even. Thus, in that case, we get that $$\int_{-1}^1P_n(x)P_m^\prime(x)\,\mathrm dx=0$$ If I call the integral $I(n,m)$ (in your notation $I(j,i)$), then so far we have that

  • $I(n,m)=0$ if $m-n$ is even.
  • $I(n,m)=0$ if $m\leq n$.

Suppose $m=n+1$, then $P_{n+1}^{(n+1)}(x)$ is simply the coefficient of $x^{n+1}$ times $(n+1)!$, which is $\frac{(2n+2)!}{2^{n+1}(n+1)!}$. Thus, $$I(n,n+1)=\frac{(2n+2)!}{2^{2n+1}n!(n+1)!}\int_{-1}^1(1-x^2)^n\,\mathrm dx$$ The integral is standard (see here). Thus, after easy cancellations, we obtain $$I(n,n+1)=2$$ So far,

  • $I(n,m)=0$ if $m-n$ is even.
  • $I(n,m)=0$ if $m\leq n$.
  • $I(n,n+1)=2$.
$\endgroup$
1
  • 1
    $\begingroup$ Based on your conclusion, I have derived the remaining part: What remains is to prove $I(n,n+1+2k) = 2, \ k\geq 1$ . By $$ \frac{d}{dx}P_{n+1}(x) = (n+1)P_n(x) + x \frac{d}{dx}P_n(x). $$ we obtain $$ \begin{align} I(n,n+1+2k) & = \int P_{n}((n+1+2k)P_{n+2k}+xP_{n+2k}) \, dx \\ & = \int xP_{n}P'_{n+2k} \, dx \\ & = xP_{n}P_{n+2k}|_{-1}^1 - \int P_{n+2k}(P_{n}+xP_{n}') \, dx \\ \end{align} $$ By the fact that $P_n(1) = 1, P_n(-1) = (-1)^n$ and that $xP_n'$ can be expressed by $P_{0},\dots,P_{n}$, it is esay to obtain $$ I(n,n+1+2k) = 2. $$ Thank you! $\endgroup$
    – luyipao
    Commented Apr 21 at 12:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .