Is this identity I found playing around with generating function with coefficients $I(n) = \int_{0}^\pi \sin^n(x) dx$ useful and or reducible?

Question

Let $I(n) = \int_{0}^\pi \sin^n(x) dx$ , using $\sin^2(x) = 1-\cos^2(x)$ and integrating by parts we get.

$$ \begin{align} I(n) = \dfrac{n-1}{n} I(n-2) \end{align} $$

With $I(0) = \pi$ and $I(1) = 2$ now let $y: \mathbb{R} \to \mathbb{R}$ be the generating function with coefficients $I(n)$ e.g. $$y(x) = I(0) +I(1)x + I(2) x^2 +\cdots$$ Since $nI(n) -(n-1) I(n-2) = 0$ we have, $$ y'(1-x^2) - xy = 2$$ subject to $y(0) = \pi$ and $y'(0) = 2$

This ODE has the solution: $$ y(x) = 2\dfrac{\arcsin(x)}{\sqrt{1-x^2}} + \dfrac{\pi}{\sqrt{1-x^2}} $$ This implies $$ \dfrac{d}{dx} \int y(x) dx = \dfrac{d}{dx} \left[ \arcsin^2(x) +\pi \arcsin(x) \right]$$ since the Taylor series for $\arcsin(x)$ is $$\sum_{n=0}^\infty \dfrac{x^{2n+1}}{2^{2n} (2n+1)} {2n \choose n} $$ so combining the above we get $$ y(x) = 2\sum_{n=0}^\infty\sum_{m=0}^\infty \dfrac{x^{2n + 2m +1}}{2^{2n+2m} (2n+1)} {2n \choose n}{2m \choose m} + \pi \sum_{n=0}^\infty \dfrac{x^{2n}}{2^{2n}} {2n \choose n} $$ We can rewrite the expression as, $$y(x) = 2\sum_{n=0}^\infty \dfrac{x^{2n+1}}{2^{2n}} \sum_{m=0}^n \dfrac{{2m \choose m} {2n-2m \choose n-m}}{2m+1} + \pi \sum_{n=0}^\infty \dfrac{x^{2n}}{2^{2n}} {2n \choose n}$$ finally by our assumption that the coefficient of $x^n$ is $I(n)$ and using the fact that: $$I(2n+1) = 2\prod_{m =1}^n \dfrac{2m}{2m+1}$$ and $$I(2n) = \pi \prod_{m=1}^n \dfrac{2m-1}{2m}$$ we get $$\prod_{m =1}^n \dfrac{2m}{2m+1} = \dfrac{1}{2^{2n}} \sum_{m=0}^n \dfrac{{2m \choose m} {2n-2m \choose n-m}}{2m+1} $$

This expression just looks funky. I plugged in some values and it does seem to hold. I can't seem to find anything about this expression but maybe it just isn't that useful?

Answer 1

It certainly gives rise to an interesting proof. Suppose we claim that

$$\frac{2^{4n}}{2n+1} {2n\choose n}^{-1} = \sum_{m=0}^n \frac{1}{2m+1} {2m\choose m} {2n-2m\choose n-m}.$$

Re-indexing we find,

$$\sum_{m=0}^n \frac{1}{2n-2m+1} {2m\choose m} {2n-2m\choose n-m} \\ = [z^{2n+1}] \log\frac{1}{1-z} \sum_{m=0}^n z^{2m} {2m\choose m} {2n-2m\choose n-m} \\ = [z^{2n+1}] \log\frac{1}{1-z} \sum_{m=0}^n z^{2m} {2m\choose m} [w^{n-m}] \frac{1}{\sqrt{1-4w}}.$$

We see that the extractor in $w$ enforces the upper range of the sum,

$$[z^{2n+1}] \log\frac{1}{1-z} [w^n] \frac{1}{\sqrt{1-4w}} \sum_{m\ge 0} z^{2m} {2m\choose m} w^m \\ = [z^{2n+1}] \log\frac{1}{1-z} [w^n] \frac{1}{\sqrt{1-4w}} \frac{1}{\sqrt{1-4wz^2}}.$$

Working with the square roots,

$$[w^n] \frac{1}{\sqrt{1-4w}} \frac{1}{\sqrt{1-4w-4w(z^2-1)}} \\ = [w^n] \frac{1}{1-4w} \frac{1}{\sqrt{1-4w(z^2-1)/(1-4w)}} \\ = \sum_{q=0}^n 4^{n-q} [w^q] \sum_{p=0}^q {2p\choose p} \frac{w^p (z^2-1)^p}{(1-4w)^p} \\ = \sum_{q=0}^n 4^{n-q} \sum_{p=0}^q {2p\choose p} (z^2-1)^p 4^{q-p} {q-1\choose p-1} \\ = 4^n \sum_{q=0}^n \sum_{p=0}^q {2p\choose p} (z^2-1)^p 4^{-p} {q-1\choose p-1} \\ = 4^n \sum_{p=0}^n {2p\choose p} (z^2-1)^p 4^{-p} \sum_{q=p}^n {q-1\choose p-1} \\ = 4^n \sum_{p=0}^n {2p\choose p} (z^2-1)^p 4^{-p} \sum_{q=0}^{n-p} {q+p-1\choose p-1} \\ = 4^n \sum_{p=0}^n {2p\choose p} (z^2-1)^p 4^{-p} [v^{n-p}] \frac{1}{1-v} \frac{1}{(1-v)^p} \\ = 4^n \sum_{p=0}^n {2p\choose p} (z^2-1)^p 4^{-p} {n\choose p} \\ = 4^n \sum_{p=0}^n {2p\choose p} 4^{-p} {n\choose p} \sum_{q=0}^p {p\choose q} (-1)^{p-q} z^{2q}.$$

Applying the extractor in $z$,

$$4^n \sum_{p=0}^n {2p\choose p} 4^{-p} {n\choose p} \sum_{q=0}^p {p\choose q} (-1)^{p-q} \frac{1}{2n+1-2q}.$$

For the inner sum introduce

$$f(z) = \frac{p!}{2n+1-2z} \prod_{r=0}^p \frac{1}{z-r}.$$

This has the property that

$$\;\underset{z=q}{\mathrm{res}}\; f(z) = \frac{p!}{2n+1-2q} \prod_{r=0}^{q-1} \frac{1}{q-r} \prod_{r=q+1}^p \frac{1}{q-r} \\ = \frac{p!}{2n+1-2q} \frac{1}{q!} \frac{(-1)^{p-q}}{(p-q)!} = {p\choose q} (-1)^{p-q} \frac{1}{2n+1-2q}.$$

Residues sum to zero and the residue at infinity is zero by inspection so we may evaluate the sum using minus the residue at $z=n+1/2$ writing

$$f(z) = - \frac{1}{2} \frac{p!}{z-(n+1/2)} \prod_{r=0}^p \frac{1}{z-r}$$

to get

$$\frac{1}{2} p! \prod_{r=0}^p \frac{1}{n+1/2-r} = \frac{1}{2} p! \frac{1}{(p+1)!} {n+1/2\choose p+1}^{-1} = \frac{1}{2n+1} {n-1/2\choose p}^{-1}.$$

Collecting what we have,

$$\frac{4^n}{2n+1} \sum_{p=0}^n {2p\choose p} 4^{-p} {n\choose p} {n-1/2\choose p}^{-1}.$$

For the quotient of the right binomial coefficients we find

$$\prod_{r=0}^{p-1} \frac{n-r}{n-1/2-r} = \prod_{r=0}^{p-1} \frac{2n-2r}{2n-1-2r} \\ = 2^p \frac{n!}{(n-p)!} \prod_{r=0}^{p-1} \frac{1}{2n-1-2r} \\ = 2^p \frac{n!}{(n-p)!} \frac{(2n-1-2p)!}{(2n-1)!} \frac{2^{n-1}(n-1)!}{2^{n-p-1} (n-1-p)!} \\ = 2^{2p} {2n-1\choose n}^{-1} {2n-1-2p\choose n-1-p}.$$

This was for $p\lt n$. Note that the above yields $2^{2n} {2n-1\choose n}^{-1}$ for $p=n$ but we should get

$${n-1/2\choose n}^{-1} = \frac{n!}{\prod_{r=0}^{n-1} (n-1/2-r)} = \frac{n! 2^n}{\prod_{r=0}^{n-1} (2n-1-2r)} \\ = n! 2^n \frac{2^{n-1} (n-1)!}{(2n-1)!} = 2^{2n-1} {2n-1\choose n}^{-1}.$$

To merge the two cases we use ${2n-1-2p\choose n-1-p} = \frac{1}{2} {2n-2p\choose n-p}$ to get

$$2^{2p-1} {2n-1\choose n}^{-1} {2n-2p\choose n-p}.$$

We now obtain for $p=n$ the value $2^{2n-1} {2n-1\choose n}^{-1}$ as required. Re-capitulating what we have,

$$\frac{4^n}{2n+1} {2n\choose n}^{-1} \sum_{p=0}^n {2p\choose p} {2n-2p\choose n-p} \\ = \frac{4^n}{2n+1} {2n\choose n}^{-1} \sum_{p=0}^n [z^p] \frac{1}{\sqrt{1-4z}} [z^{n-p}] \frac{1}{\sqrt{1-4z}} \\ = \frac{4^n}{2n+1} {2n\choose n}^{-1} [z^n] \frac{1}{1-4z} = \frac{2^{4n}}{2n+1} {2n\choose n}^{-1}.$$

This is the claim. Here have used that $\frac{1}{2} {2n-1\choose n}^{-1} = {2n\choose n}^{-1}.$

Answer 2

You can as well prove the equation directly by going in the opposite direction, although it doesn't seem to give much value or insight into what's going on or why.

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We need to show that

$$ \frac{1}{2^{2n}} \sum\limits_{m=0}^n \frac{1}{2m+1} \binom{2m}{m}\binom{2(n-m)}{n-m} = \prod\limits_{m=1}^n \frac{2m}{2m+1} = f_{2n+1} $$

Let's find the generating function

$$ G(x) = \sum\limits_{n=0}^\infty f_{2n+1} x^{2n+1} $$

from the LHS:

$$ 2\sum\limits_{n=0}^\infty \sum\limits_{m=0}^n \left[\frac{(x/2)^{2m+1}}{2m+1} \binom{2m}{m}\right]\left[ \binom{2(n-m)}{n-m} (x/2)^{2(n-m)}\right] = 2 \left[\int A(\frac{x}{2})d\frac{x}{2}\right]\left[A(\frac{x}{2})\right], $$ where $A(x) = \sum\limits_{m=0}^\infty x^{2m} \binom{2m}{m} = \frac{1}{\sqrt{1-4x^2}}$, from which we get

$$ \boxed{G(x) = \frac{\arcsin x}{\sqrt{1-x^2}}} $$

This is essentially the odd part of the generating function that the OP got from the diffeq.

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Now, from the right hand side, we get

$$ f_{n} = \frac{n-1}{n} f_{n-2} $$

with the base cases $f_{2n} = f_0=0$ and $f_1 = 1$. Removing denominators, we get

$$ n f_n = (n-1) f_{n-2}, $$

which in terms of the generating function

$$ F(x) = \sum\limits_{n=0}^\infty f_{2n+1} x^{2n+1} $$

and using the notation $\Delta = x \frac{\partial}{\partial x}$, translates into a diffeq:

$$ \boxed{\Delta F(x) = (\Delta - 1) x^2 F(x) + x} $$

Or, equivalently in terms of derivatives:

$$ (1-x^2)F'(x)= x F(x) + 1. $$

Surely enough, the solution is $F(x) = \frac{c_1+\arcsin x}{\sqrt{1-x^2}}$, and $c_1=0$ corresponds to $f_0=0$.

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I hope someone eventually provides a meaningful combinatorial/probabilistic interpretation of this...