1
$\begingroup$

I have a generating function of Legendre Polynomials given by: $G(x,r)= \sum_{n=0}^\infty P_n(x)r^n = (1-2rx +r^2)^{-1/2}$

My problem is that I'm asked to find $\int_{-1}^1P_n(x)dx$ but all I have is $P_n(x)r^n$, I'm not sure how to (essentially) remove the $r^n$ so that I have what I need.

I can work out the integral myself I would just like to know how to get from $P_n(x)r^n$ to $P_n(x)$.

One method I thought of was to define some $C_n$ so that:

$\int_{-1}^1P_n(x)dx=C_n$

Then I could integrate both sides but I got confused as to what would be in the $(...)$

$\sum_n C_nr^n=\int_{-1}^1(...)dx$

Any help would be much appreciated

$\endgroup$

1 Answer 1

2
$\begingroup$

Since $$\int_{-1}^1 G(x,r) \;dx = 2$$ it follows that $$\int_{-1}^1 P_n(x)\;dx = \begin{cases} 2 \qquad \text{if } n = 0 \\ 0 \qquad \text{otherwise} \end{cases}$$

$\endgroup$
4
  • $\begingroup$ Oh, I didn't realise it was that simple. Did you get that by using the fact that both $|x| <=1 $ and $|r| <=1$? Or is it a general rule that the integration between 1 and -1 of a generating function is equal to 2? $\endgroup$
    – Charlie P
    Commented May 18, 2021 at 14:15
  • $\begingroup$ Or did you just integrate $G(x,r)$? When i Integrate $G(x,r)dx$ I get $2/r$, in my question it says that $|r|<1$ (I made a mistake in my previous comment) $\endgroup$
    – Charlie P
    Commented May 18, 2021 at 14:21
  • $\begingroup$ @CharlieP Just integrate $G(x,r)$ with respect to $x$ from $-1$ to $1$. The result is $2$, with no dependency on $r$. $\endgroup$
    – awkward
    Commented May 18, 2021 at 14:41
  • $\begingroup$ Yep, I realised it after I sent the comment that you integrated $G(x,r)$ I just had a small-brain moment. Thanks for the help $\endgroup$
    – Charlie P
    Commented May 18, 2021 at 14:49

You must log in to answer this question.

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