Showing a summation identity for $1$, possibly tied to Legendre polynomials

Question

The Problem: Consider the sign function on $(-1,0)\cup(0,1)$ defined by

$$ \sigma(x) := \left. \text{sgn}(x) \right|_{(-1,0)\cup(0,1)} = \begin{cases} 1 & x \in (0,1) \\ -1 & x \in (-1,0) \end{cases}$$

The problem is to show that

$$\int_{-1}^1 (\sigma(x))^2 dx = 2 \sum_{n=0}^\infty (4n+3) \left( \frac{(2n-1)!!}{(2n+2)!!} \right)^2$$

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Context: This is (in its essence) problem $15.2.8$ in Mathematical Methods for Physicists by Arfken, Weber, & Harris. It was assigned to me as a homework problem for one of my classes. (In that vein I would prefer to only have nudges in the right direction, rather than full solutions.) The discussion in this section ($\S 15.2$) is essentially on Legendre polynomials and Fourier-Legendre series.

It is quite obvious that the integral evaluates to $2$, so the problem is ultimately showing that

$$\sum_{n=0}^\infty (4n+3) \left( \frac{(2n-1)!!}{(2n+2)!!} \right)^2 = 1$$

However, browsing the text, Wikipedia, and MathWorld don't give me any enlightening ideas on what identities to leverage. Expanding $f(x) = 1$ as a Fourier-Legendre series

$$f(x) = \sum_{n=0}^\infty c_n P_n(x) \; \text{where} \; c_n = \int_{-1}^1 f(x)P_n(x)dx$$

doesn't really lead me anywhere (for the integral in $c_n$ is zero whenever $n \ge 1$) - which is obvious enough, since $P_0(x) = 1$ anyways, so of course we'd get a finite series.

The identity does seem true. Taking the equivalent formulation of the problem (as a series equaling $1$) and summing $n=0$ to $n=100$ yields a result of about $0.996$ according to Wolfram, and up to $n=10,000$ yields about $0.999354$ (Wolfram), so it seems reasonable it converges to $1$, albeit somewhat slowly.

The original problem is in multiple parts: this is part (a), and part (c) notes, as I did, the integral $\int_{-1}^1 \sigma^2(x)dx = 2$. So it also seems plausible that I'm not even meant to calculate the integral at the outset, but instead utilize some other method. I suppose one could rewrite $\sigma$ as

$$ \sigma(x) = \begin{cases} P_0(x) & x \in (0,1) \\ -P_0(x) & x \in (-1,0) \end{cases}$$

and perhaps utilize some sort of identity used in the motivations/derivations tied to Legendre polynomials (a lot of integrals of $P_n^2$ seem to come up), but this rewriting doesn't give me anything more enlightening to work with.

Does anyone have some ideas as to how I might at least get started with this?

Answer 1

Thanks to @metamorphy and his insights about similiarity to Parseval's theorem, the answer is achieved, in essence, by expanding $\sigma$ as a Fourier-Legendre series, which I detail below. (Not in its full detail, though, mostly recollecting the high points.)

$\newcommand{\dd}{\mathrm{d}}$ $\newcommand{\para}[1]{\left( #1 \right)}$ $\newcommand{\encla}[1]{\langle #1 \rangle}$

We begin by first expanding $\sigma$ as a Fourier-Legendre series; recall this takes the form $$ \sigma(x) = \sum_{n=0}^\infty c_n P_n(x) \text{ where } c_n = \frac{2n+1}{2} \int_{-1}^1 \sigma(x)P_n(x) \, \dd x $$ We begin calculating the constants $c_n$. Note that, due to the piecewise nature of $\sigma$, $$ c_n = \frac{2n+1}{2} \para{ \int_{-1}^0 -P_n(x) \, \dd x + \int_0^1 P_n(x) \, \dd x } $$ We utilize that $P_n$ is even if $n$ is even here, and similar for $n$ odd. This will lead us to conclude $c_n = 0$ for $n$ even, and for $n$ odd, $$ c_{\text{n, odd}} = (2n+1) \int_0^1 P_n(x) \, \dd x $$ To calculate what remains, consider the recursion $$ \frac{P_{n+1}' (x) - P_{n-1}'(x)}{2n+1} = P_n(x) $$ Integrate both sides of this over $x \in (0,1)$. Then it readily follows that $$ \int_0^1 P_n(x) \, \dd x =\left. \frac{1}{2n+1} \Big( P_{n+1}(x) - P_{n-1}(x) \Big) \right|_{x=0}^1 $$ Due to the normalization process, $P_n(1) = 1$ always. Meanwhile, $$ P_n(0) = \begin{cases} (-1)^{n/2} \frac{(n-1)!!}{n!!} & n \equiv 0 \pmod 2 \\ 0 & n \equiv 1 \pmod 2 \end{cases} $$ Using this, the assumption $n$ is odd, and a multitude of algebraic manipulations (mainly factoring and properties of double factorials) we conclude that, if $n = 2k+1$, $$ \int_0^1 P_{2k+1} \, \dd x = (-1)^k \frac{(2k-1)!!}{(2k+2)!!} $$ Thus, if $n$ is even, $c_n = 0$; if $n = 2k+1$ is odd, then $$ c_{2k+1} = (4k+3) (-1)^k \frac{(2k-1)!!}{(2k+2)!!} $$ This means that we can simply sum over the odd indices in our series for $\sigma$, switch our dummy variable from $k$ to $n$, and conclude $$ \sigma(x) = \sum_{n=0}^\infty (4n+3) (-1)^n \frac{(2n-1)!!}{(2n+2)!!} P_{2n+1}(x) $$ We then choose to square this representation of $\sigma$, using the Cauchy product: $$ \sigma^2(x) = \left( \sum_{n=0}^\infty c_n P_n(x) \right)^2 = \sum_{n=0}^\infty \sum_{m=0}^n c_{2n+1} c_{2(n-m)+1} P_{2n+1}(x) P_{2(n-m)+1}(x) $$ We now integrate throughout with respect to $x \in (-1,1)$, and assume there is no issue in interchanging summation and integration here. Then we see $$ \int_{-1}^1 \sigma^2(x) \, \dd x = \sum_{n=0}^\infty \sum_{m=0}^n c_{2n+1} c_{2(n-m)+1} \int_{-1}^1 P_{2n+1}(x) P_{2(n-m)+1}(x) \, \dd x $$ We recall the orthogonality relation, $\encla{P_p,P_q} = \frac{2}{2p+1} \delta_{p,q}$. This causes much simplification, yielding $$ \int_{-1}^1 \sigma^2(x) \, \dd x = \sum_{n=0}^\infty c_{2n+1}^2 \frac{2}{4n+3} $$ We bring back in our values for the $c_{2n+1}$, and square them. A factor of $4n+3$ cancels in this process. We can then bring the $2$ outside of the sum, to conclude with our desired result: $$ \int_{-1}^1 \sigma^2(x) \; \dd x = 2 \sum_{n=0}^\infty (4n+3) \para{ \frac{(2n-1)!!}{(2n+2)!!} }^2 $$