Contour integral with power of sine

Question

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.

$$\lim_{R \to \infty} \int_{0}^{\pi/2\alpha} \exp(i R^\alpha \exp(i \alpha \theta)) R i \exp(i \theta) d\theta = 0$$

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}\sin\pars{x^{\alpha}}\dd x \,\right\vert_{\,\alpha\ >\ 1} = \sin\pars{\pi \over 2\alpha} \int_{0}^{\infty} \exp(-x^{\alpha})\,\dd x}:\ {\Large ?}}$.

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\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}