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

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

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

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?

• The appearance of the problem begs to consider the F-L expansion of $\sigma(x)$. – metamorphy Oct 3 '20 at 6:43
• Hm. I suppose I can try that tomorrow, though the motivation doesn't feel clear. Why do you think that this situation, to put it as you did, "begs" for it? Clearly I'm missing something obvious if so. – Eevee Trainer Oct 3 '20 at 8:37
• Because of the way the integral is written (Parseval's identity). – metamorphy Oct 3 '20 at 8:39
• Oh, okay, I see now. I hadn't used that in a few years so I totally forgot about that. Thanks for the insight, I'll see what this results in! – Eevee Trainer Oct 3 '20 at 21:53

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.)
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: $$\int_{-1}^1 \sigma^2(x) \; \dd x = 2 \sum_{n=0}^\infty (4n+3) \para{ \frac{(2n-1)!!}{(2n+2)!!} }^2$$