Very interesting integral limit I found this interesting problem on AoPS forum but no one has posted an answer. I have no idea how to solve it.

$$
\int_0^\infty \sin(x^n)\,dx
$$
  For all positive rationals $n>1$, $I_n$ denotes the integral as above.
If $P_n$ denotes the product
  $$
P_n=\prod_{r=1}^{n-1}I_{\bigl(\!\frac{n}{r}\!\bigr)}\,,
$$
  then evaluate the following limit $L$
  $$
L=\lim_{n\to\infty}\bigl(\sqrt{n}\,P_n\bigr)^{\frac{1}{n}}
$$

 A: since
$$T_{\frac{r}{n}}=\int_{0}^{+\infty}\sin{(x^{\frac{r}{n}})}dx=\Gamma\left(\dfrac{r}{n}+1\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}},r=1,2,,\cdots,n-1$$
proof see:3.1
Lemma $1$:
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}=\dfrac{\sqrt{n}}{2^{n-1}}$$
Use this well known
$$z^{n-1}+z^{n-2}+\cdots+z+1=\prod_{k=1}^{n-1}(z-(z-z_{k})$$
let $z=1$ then we have
$$n=\prod_{i=1}^{n-1}\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|$$
since
$$\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|=2\sin{\dfrac{k\pi}{n}}$$
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{n}}=\dfrac{n}{2^{n-1}}$$
and
$$\dfrac{2n}{2^{2n-1}}=\prod_{i=1}^{2n-1}\sin{\dfrac{i\pi}{2n}}=\left(\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}\right)^2$$
By done!
Lemma 2:   $$f=\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{2}{n}\right)\cdots\Gamma\left(\dfrac{n-1}{n}\right)=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$
since
$$f^2=\left(\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{n-1}{n}\right)\right)\cdot \left(\Gamma\left(\dfrac{2}{n}\right)\Gamma\left(\dfrac{n-2}{n}\right)\right)\cdots\left(\Gamma\left(\dfrac{n-1}{n}\right)\Gamma\left(\dfrac{1}{n}\right)\right)=\dfrac{\pi}{\sin{\frac{\pi}{n}}}\cdot\dfrac{2\pi}{\sin{\frac{2\pi}{n}}}\cdots\dfrac{\pi}{\sin{\dfrac{(n-1)\pi}{n}}}$$
so we have 
$$f=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$
so we have
$$P_{n}=\prod_{r=1}^{n-1}T_{\frac{r}{n}}=\prod_{r=1}^{n-1}\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{(n-1)!}{n^{n-1}}\cdot\dfrac{\sqrt{n}}{2^{n-1}}\cdot \dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}=\left(\dfrac{\sqrt{2\pi}}{2n}\right)^{n-1}(n-1)!$$
By stirling formula 
 we have
$$\lim_{n\to+\infty}(\sqrt{n}P_{n})^{\frac{1}{n}}=\dfrac{\sqrt{2\pi}}{2e}$$
A: The integral $I(a)=\int_0^{\infty}\sin(x^a)=\Im\int_0^{\infty}\exp(ix^a)$ is easily calulated by using the analyticality of the integrand: Rotate the the contour of integration by an angle of $\frac{\pi}{2a}$ to get $I(a)=\Im\left(e^{\frac{i\pi}{2a}}\int_0^{\infty}e^{-x^a} \right)$ or 

$$
I(a)=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\sin\left(\frac{\pi}{2a}\right)
$$

Now let us have a look at 
$$
P_n=\left(n^{\frac12}\prod_{r=1}^{n-1}I\left(\frac{n}r\right)\right)^{\frac1n}=n^{\frac1{2n}}\exp\left(\frac1n\underbrace{\sum_{r=1}^{n-1}\log\left(I\left(\frac{n}{r}\right)\right)}_{s_{n-1}}\right)
$$
Now we observe that $I(1)=0$ so $s_n=s_{n-1}$ so 
$$
\frac{1}{n}s_{n-1}=\frac{1}{n}s_n=\frac{1}{n}\sum_{r=1}^{n}\log\left(\frac{r}{n}\sin\left(\frac{\pi r}{2n}\right)\Gamma\left(\frac{r}{n}\right)\right)
$$
which is the defintion of a Riemannian integral. Since also the prefactor in $P_n$ is finite ($n^{\frac{1}{2n}}\sim_{\infty}1$) we can split the limit

$$
L=\lim_{n\rightarrow\infty}P_n=\lim_{n\rightarrow\infty}n^{\frac{1}{2n}}\lim_{n\rightarrow\infty}e^{\frac{1}{n}s_n}=1\cdot e^J \,\,\quad \left(\star\right)
$$

where $J$ is given by
$$
J=\int_0^1\log(I(x))dx=\\\int_0^1\log(x)dx+\int_0^1\log\left(\sin\left(\frac{\pi }{2}x\right)\right)dx+\int_0^1\log\left(\Gamma(x)\right)dx
$$
or 
$$
J=-1-\log(2)+\frac{1}{2}\log(2\pi)
$$
where we used a classic result by Raabe as well as this all time favourite. 
We now can conclude from $\left(\star\right)$ that

$$
L=\frac{\sqrt{2\pi}}{2e}
$$

in accordance with @math110
