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i want to evaluate the following integral:

\begin{equation} a_n = \frac{2n+1}{2}\int_{-1}^1 \frac{P_n(x)}{\sqrt{2-2x}} \text{dx} \ \text{where $P_n(x)$ is the $n^{th}$ Legendre Polynomial. } \end{equation} I am expecting the result to be $1$. I tried to set it up on MAPLE and tried a large number of $n$'s ( 1,2 ,3,1000,2000) and the result is $1$. However, i am looking for a rigorous proof. I looked into the table of integrals involving legendre polynomials and could not find a case that suits mine.

Can anyone help me please?

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2 Answers 2

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The generating function for Legendre polynomials is $$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n}P_n(x)t^n$$

Just use $t=1$ in here to expand your integrand in Legendre polynomials and then use the orthogonality condition.

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  • $\begingroup$ Can you provide me the reference where we are allowed to replace $t$ by 1. All the references state that $|t|$ < 1. $\endgroup$
    – outlaw
    Oct 11, 2017 at 16:25
  • $\begingroup$ Why on earth was this downvoted? $\endgroup$
    – Marcel
    Oct 23, 2017 at 10:53
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    $\begingroup$ @Marcel: Probably because you do not state on what subsets that equality is valid (i.e. what are the allowed values for $x$ and $t$ in it). In particular, how does one know whether the power series in $t$ converges in $t=1$? What happens for $t=x=1$? $\endgroup$
    – Alex M.
    Nov 4, 2017 at 16:21
  • $\begingroup$ @AlexM. This is not intended as general theory, but only as an answer to the question. $\endgroup$
    – Marcel
    Nov 4, 2017 at 18:28
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    $\begingroup$ @Marcel: First, I'm not the downvoter. Second, even as an answer to the question it shouldn't ignore studying the matters of convergence that arise (and I see two of them: the convergence of the series and the convergence of some improper integrals at $1$). $\endgroup$
    – Alex M.
    Nov 4, 2017 at 19:28
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Use the following identity for $P_n$ as a sum of polynomials

\begin{equation} P_n(x) = \sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}\left(\frac{x-1}{2}\right)^k. \end{equation}

Since this is a finite sum there's no convergence issues plugging into the integral and bringing out the summation. This will give us

\begin{align*} a_n &= \frac{2n+1}{2\sqrt{2}}\int_{-1}^1 \frac{P_n(x)}{(1-x)^{1/2}}\;dx\\ &= \frac{2n+1}{2\sqrt{2}}\int_{-1}^1 \frac{\sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}(1-x)^{k}(-1)^{k}2^{-k}}{(1-x)^{1/2}}\;dx\\ &= \frac{2n+1}{2\sqrt{2}}\sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}(-1)^{k}2^{-k}\int_{-1}^1 (1-x)^{k-\frac{1}{2}} \;dx\\ &= \frac{2n+1}{2\sqrt{2}}\sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}(-1)^{k}2^{-k}\left(\frac{2^{k+\frac{1}{2}}}{k+\frac{1}{2}}\right)\\ &= \frac{2n+1}{2}\sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}(-1)^{k}\left(\frac{1}{k+\frac{1}{2}}\right)\\ \end{align*}

A quick check of the sum via Wolfram Alpha shows that

\begin{equation} \sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k}(-1)^{k}\left(\frac{1}{k+\frac{1}{2}}\right) = \frac{2}{2n+1}. \end{equation}

Whence $a_n=1$.

Depending on how rigorous of a proof you want, I'll leave it to you to derive the summation above via binomial identities.

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