\(\displaystyle \displaystyle \int_{0}^{\infty}sin(x^{2n})dx=sin(\frac{\pi}{4n})\int_{0}^{\infty}e^{-x^{2n}}dx\)
Let \(\displaystyle 2n=p\)
\(\displaystyle \displaystyle \int_{0}^{\infty}sin(x^{p})dx\)
Then, using imaginary parts:
\(\displaystyle -\displaystyle \int_{0}^{\infty}e^{-ix^{p}}dx\)
Let \(\displaystyle x=t\cdot i^{-1/p}, \;\ dx=i^{-1/p}dt=-sin(\frac{\pi}{2p})\)
\(\displaystyle -i^{-1/p}\displaystyle \int_{0}^{\infty}e^{-t^{p}}dt\)
Let \(\displaystyle u=t^{p}\)
\(\displaystyle \frac{-1}{p} i^{-1/p}\displaystyle \int_{0}^{\infty}u^{\frac{1}{p}-1}e^{-u}du\)
\(\displaystyle =\frac{1}{p}\Gamma(1/p)sin(\frac{\pi}{2p})\)
\(\displaystyle =\Gamma(1+\frac{1}{p})sin(\frac{\pi}{2p})\)
Now, resub \(\displaystyle p=2n\) and we get:
\(\displaystyle sin(\frac{\pi}{2n})\Gamma(1+\frac{1}{2n})\)
Now, use the fact that \(\displaystyle \displaystyle \int_{0}^{\infty}e^{-x^{2n}}dx=\Gamma(1+\frac{1}{2n})\)
This last identity comes from an alternative defintion of Gamma:
\(\displaystyle \Gamma(n)=\displaystyle 2\int_{0}^{\infty}x^{2n-1}e^{-x^{2}}dx\)
