Almost Walli's Integrals

An outline that shows the main steps: We have that $\int_0^\infty \sin(x^n) \: dx = \sin \left( \frac{\pi}{2n} \right) \Gamma \left( \frac{n + 1}{n} \right).$ (For example, see http://math.stackexchange.com/questions/453811/integral-of-int-limits-0-infty-frac-sin-xn-xndx .) Then $\prod_{r = 1}^{n - 1} I_{n/r} = \prod_{r = 1}^{n - 1} \sin \frac{\pi r}{2n} \cdot \prod_{r = 1}^{n - 1} \Gamma \left( \frac{n + r}{n} \right).$

It can be shown that $\prod_{r = 1}^{n - 1} \sin \frac{\pi r}{2n} = \frac{\sqrt{n}}{2^{n - 1}}$ and $\prod_{r = 1}^{n - 1} \Gamma \left( \frac{n + r}{n} \right) = (2 \pi)^{(n - 1)/2} \cdot n^{1/2 - n} \cdot (n - 1)!,$ so $L = \lim_{n \to \infty} \left( \frac{n}{2^{n - 1}} \cdot (2 \pi)^{(n - 1)/2} \cdot n^{1/2 - n} \cdot (n - 1)! \right)^{1/n}.$ This works out to $L = \frac{\sqrt{2 \pi}}{2e}.$

Solving the integral $\displaystyle\int\limits_0^\infty \sin(x^n) dx = \sin(\frac{\pi}{2n})\gamma(\frac{n+1}{n})$

Given $P_n=\prod_{r=1}^ {n-1} I(\frac{n}{r})$

Solving the limit:

Let $y=\lim_{n \to \infty} \left(\sqrt{n}P_n\right)^{1/n}$

$log(y)=\lim_{n \to \infty} \frac{1}{n} log{\sqrt{n}P_n}$

$log(y)=\lim_{n \to \infty} \frac{log\sqrt{n}}{n} + \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n-1} log(\sin(\frac{\pi r}{2n})\gamma(1+\frac{r}{n})$

$log(y)=\int_0^1 log(\sin(\frac{\pi x}{2})\gamma(1+x)) dx$

I get

$log(y)= -0.7742086$

$y=0.461068$

$1000y=461.068$

The greatest integer value is $461$ , which is the required answer.

I will share my solution on how

$\displaystyle \int_{0}^{\infty}{sin({x}^{n}) \ dx} = sin\left(\frac{\pi}{2n}\right)\Gamma\left(\frac{n+1}{n}\right)$

Okay, let's start.

First of all, we must notice some things that since,

$\displaystyle sin(x) = \sum_{k=0}^{\infty}{\frac{{(-1)}^{k} {x}^{2k+1}}{(2k+1)!}}$

Then, $\displaystyle \frac{sin(\sqrt{x})}{\sqrt{x}} = \sum_{k=0}^{\infty}{\frac{{(-x)}^{k}}{(2k+1)!}} = \sum_{k=0}^{\infty}{\frac{\Gamma(k+1)}{\Gamma(2k+2)} \frac{{(-x)}^{k}}{k!}}$

Well, that is the exponential generating function and it might be obvious now that I will use Ramanujan Master Theorem.

$\displaystyle \int_{0}^{\infty}{sin({x}^{n}) \ dx} = \frac{1}{2n} \int_{0}^{\infty}{\frac{sin(\sqrt{u})}{\sqrt{u}} {u}^{\frac{n+1}{2n} -1} \ du}$ [As we substitute $\sqrt{u} = {x}^{n}$ ]

$\displaystyle = \frac{1}{2n} M\left(\frac{sin(\sqrt{u})}{\sqrt{u}} ; \frac{n+1}{2n}\right)$ where $M(f ; {a}_{n})$ means Mellin Transform of $f$ at ${a}_{n}$ .

Now, using Ramanujan Master Theorem,

$\displaystyle = \frac{1}{2n} \Gamma\left(\frac{n+1}{2n}\right) \frac{\Gamma\left(1 - \frac{n+1}{2n}\right)}{\Gamma\left(2 - 2\frac{n+1}{2n}\right)}$

$\displaystyle = \frac{1}{2n} \frac{\pi}{sin\left(\frac{\pi(n+1)}{2n}\right)}\frac{1}{\Gamma\left(\frac{n-1}{n}\right)}$

$\displaystyle = \frac{1}{2n} \frac{\pi}{cos\left(\frac{\pi}{2n}\right)} \frac{\Gamma\left(\frac{1}{n}\right)sin\left(\frac{\pi}{n}\right)}{\pi}$

$\displaystyle = \frac{\Gamma\left(\frac{1}{n}\right) \ sin\left(\frac{\pi}{2n}\right)}{n}$

Using basic definition of Gamma function (factorial),

$\displaystyle = \Gamma\left(\frac{n+1}{n}\right) \ sin\left(\frac{\pi}{2n}\right)$