Contour integral with power of sine I am trying to show that, for $\alpha > 1$,
$$\int_{0}^{\infty} \sin(x^{\alpha}) dx = \sin \left ( \frac{\pi}{2\alpha} \right ) \int_{0}^{\infty} \exp(-x^{\alpha}) dx$$
The approach I've used has been to take the imaginary part of the integral of $\exp(i x^{\alpha})$ over the line. Then I'm trying to compute this with a wedge contour with angle $\pi/2\alpha$. The horizontal segment and the diagonal segment are fine, but I'm having trouble showing that the arc integral goes to zero, i.e.
$$\lim_{R \to \infty} \int_{0}^{\pi/2\alpha} \exp(i R^\alpha \exp(i \alpha \theta)) R i \exp(i \theta) d\theta = 0$$
 A: Write the integral as the imaginary part of 
$$\int_0^{\infty} dx \, e^{i x^{\alpha}}$$
Then follow this derivation to get that
$$\int_0^{\infty} dx \, e^{i x^{\alpha}} = e^{i \pi/(2 \alpha)} \int_0^{\infty} dx \, e^{-x^{\alpha}}$$
Taking imaginary parts produces the stated result. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}\sin\pars{x^{\alpha}}\dd x
\,\right\vert_{\,\alpha\ >\ 1} =
\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty} \exp(-x^{\alpha})\,\dd x}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}\sin\pars{x^{\alpha}}
\dd x\,\right\vert_{\,\alpha\ >\ 1}}
\,\,\,\stackrel{x^{\alpha}\ \mapsto\ x}{=}\,\,\,
{1 \over \alpha}\,\Im\int_{0}^{\infty}
x^{1/\alpha - 1}\,\,\expo{\ic x}\dd x
\\[5mm] = &\
{1 \over \alpha}\,\Im\int_{0}^{\infty}
\ic^{1/\alpha - 1}\,\,\,y^{1/\alpha - 1}\,\,
\expo{-y}\,\ic\,\dd y\quad
\pars{\substack{\mbox{I'll "close" the integral}
\\[0.5mm] \mbox{ along a quarter circle}
\\[0.5mm] \mbox{ in the complex plane}
\\[0,5mm] \mbox{first quadrant.}}}
\\[5mm] = &\
{1 \over \alpha}\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty}y^{1/\alpha - 1}\,\,\expo{-y}\,\dd y
\\[5mm] \stackrel{y\ =\ x^{\alpha}}{=}\,\,\,&
{1 \over \alpha}\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty}\pars{x^{\alpha}}^{1/\alpha - 1}\,\,\expo{-x^{\alpha}}\pars{\alpha x^{\alpha - 1}}\dd x
\\[5mm] = &\
\bbx{\sin\pars{\pi \over 2\alpha}\int_{0}^{\infty}
\expo{-x^{\alpha}}\dd x} \\ &
\end{align}
