计算广义积分$$\int_0^{+\infty}\cos x^p {\rm d}x,\int_0^{+\infty}\sin x^p {\rm d}x, p>1$$

${\bf 解:}$

在角状域$G=\{z\in\mathbb{C}|0<{\rm Arg}z<\frac{\pi}{2p}\}$上引入辅助函数$e^{iz^p}$, 其中$z^p=|z|^pe^{ip{\rm Arg}z}$,$0<{\rm Arg}z<\frac{\pi}{2p}$, 再设$0<\rho<R<+\infty$, 以及$\gamma_\rho=\partial B(0,\rho)\cap G$,$\gamma_R=\partial B(0,R)\cap G$, 逆时针为它们的正向. 由留数定理(或$Cauchy$积分公式), 得到

\begin{equation}\label{the1}
\int_\rho^Re^{ix^p}{\rm d}x+\int\limits_{\gamma_\rho}e^{iz^p}{\rm d}z+\int^\rho_R e^{x^p}e^{i\frac{\pi}{2p}}{\rm d}x-\int\limits_{\gamma_R}e^{iz^p}{\rm d}z=0
\end{equation}

下面证明$(\ref{the1})$中的第$2$,$4$项分别在$R\rightarrow+\infty,\rho\rightarrow 0^+$时趋向$0$.

当$R\rightarrow+\infty$时注意$e^{iz^p}=e^{iR^pe^{ip\theta}}=e^{R^p\cdot i(\cos p\theta+i \sin p\theta)}= e^{R^p\cdot (-\sin p\theta+i\cos p\theta)}$, 以及当$0<x<\frac{\pi}{2}$时成立$\sin x>\frac{2x}{\pi}$, 可得

\begin{align*} | \int\limits_{\gamma_\rho} e^{iz^p} {\rm d}z |  &\leq  \int \limits_{\gamma_\rho} |{e^{iz^p}|{\rm d}z } \\ &=  \int^\frac{\pi}{2p}_0Re^{-R^p\sin p\theta}{\rm d}\theta\\ &\leq  R\int^\frac{\pi}{2p}_0e^{-R^P\frac{2p\theta}{\pi}}{\rm d}\theta  \\ &=  -\frac{\pi R}{2pR^p}e^{-R^P\frac{2p\theta}{\pi}}|^{\frac{2p\theta}{\pi}}_0 \\ &=  \frac{\pi R}{2pR^p}(1-e^{-R^p})\\ &\rightarrow 0(R\rightarrow +\infty)\end{align*}

当$\rho\rightarrow 0^+$时,

\begin{align*}|\int\limits_{\gamma_R}e^{iz^p}{\rm d}z|&=|\int^\frac{\pi}{2p}_0 e^{i\rho^pe^{ip\theta}} \rho e^{i\theta}i{\rm d}\theta|\\&\rightarrow 0(\rho\rightarrow 0^+)\end{align*}

于是可将$(\ref{the1})$化为

\begin{align*}\int_0^{+\infty} e^{ix^p}{\rm d}x &=e^{i\frac{\pi}{2p}}\int_0^{+\infty}e^{-x^p}\\ &=e^{i\frac{\pi}{2p}}\int_0^{+\infty}e^{-t}t^{\frac{1}{p}-1}{\rm d}t\\ &=\frac{1}{p}\Gamma(\frac{1}{p})e^{i\frac{\pi}{2p}}\end{align*}

故

\begin{align}\int_0^{+\infty}\cos x^p {\rm d}x &=\frac{1}{p}\Gamma(\frac{1}{p})\cos \frac{\pi}{2p}\\  \int_0^{+\infty}\sin x^p {\rm d}x &=\frac{1}{p}\Gamma(\frac{1}{p})\sin \frac{\pi}{2p}\end{align}

葛神给出了一个数学分析的做法:

\begin{align*} \int_0^\infty \sin \left( x^n\right)dx &= \frac{1}{n}\int_0^\infty x^{\frac{1}{n}-1} \sin(x) \ dx \quad (x^n \mapsto x) \\
&= \frac{1}{n \Gamma \left(
1-\frac{1}{n}\right)}\int_0^\infty \left(\int_0^\infty
u^{-\frac{1}{n}}e^{-xu}du\right) \sin(x) \ dx\\
&= \frac{1}{n
\Gamma \left( 1-\frac{1}{n}\right)} \int_0^\infty u^{-\frac{1}{n}}
\left( \int_0^\infty e^{-xu}\sin(x) \ dx\right)du\\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)} \int_0^\infty \frac{u^{-\frac{1}{n}}}{1+u^2}du \\
&=
\frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)}
\int_0^{\frac{\pi}{2}}\tan^{-\frac{1}{n}}(\theta) d\theta \quad (u=\tan
\theta) \\
&= \frac{1}{n \Gamma \left(
1-\frac{1}{n}\right)}\int_0^{\frac{\pi}{2}}\sin^{-\frac{1}{n}}(\theta)
\cos^{\frac{1}{n}}(\theta) d\theta \\
&= \frac{1}{2n \Gamma \left( 1-\frac{1}{n}\right)} \mathrm{B} \left( \frac{1-n}{2},\frac{1+n}{2}\right) \\
&=
\frac{1}{2n \Gamma \left( 1-\frac{1}{n}\right)} \Gamma \left(
\frac{n-1}{2n}\right)\Gamma \left( \frac{n+1} {2n}\right) \\
&= \frac{\sin \left( \frac{\pi}{n}\right)}{2n\cos \left( \frac{\pi}{2n}\right)}\Gamma \left( \frac{1}{n}\right) \\
&= \frac{1}{n}\sin \left(\frac{\pi }{2n} \right)\Gamma \left( \frac{1}{n}\right)\end{align*}