${f 解:}$
在角状域$G={zinmathbb{C}|0<{ m Arg}z<frac{pi}{2p}}$上引入辅助函数$e^{iz^p}$, 其中$z^p=|z|^pe^{ip{ m Arg}z}$,$0<{ m Arg}z<frac{pi}{2p}$, 再设$0< ho<R<+infty$, 以及$gamma_ ho=partial B(0, ho)cap G$,$gamma_R=partial B(0,R)cap G$, 逆时针为它们的正向. 由留数定理(或$Cauchy$积分公式), 得到
egin{equation}label{the1}
int_
ho^Re^{ix^p}{
m d}x+intlimits_{gamma_
ho}e^{iz^p}{
m d}z+int^
ho_R e^{x^p}e^{ifrac{pi}{2p}}{
m d}x-intlimits_{gamma_R}e^{iz^p}{
m d}z=0
end{equation}
下面证明$( ef{the1})$中的第$2$,$4$项分别在$R ightarrow+infty, ho ightarrow 0^+$时趋向$0$.
当$R ightarrow+infty$时注意$e^{iz^p}=e^{iR^pe^{ip heta}}=e^{R^pcdot i(cos p heta+i sin p heta)}= e^{R^pcdot (-sin p heta+icos p heta)}$, 以及当$0<x<frac{pi}{2}$时成立$sin x>frac{2x}{pi}$, 可得
egin{align*} | intlimits_{gamma_ ho} e^{iz^p} { m d}z | &leq int limits_{gamma_ ho} |{e^{iz^p}|{ m d}z } \ &= int^frac{pi}{2p}_0Re^{-R^psin p heta}{ m d} heta\ &leq Rint^frac{pi}{2p}_0e^{-R^Pfrac{2p heta}{pi}}{ m d} heta \ &= -frac{pi R}{2pR^p}e^{-R^Pfrac{2p heta}{pi}}|^{frac{2p heta}{pi}}_0 \ &= frac{pi R}{2pR^p}(1-e^{-R^p})\ & ightarrow 0(R ightarrow +infty)end{align*}
当$
ho
ightarrow 0^+$时,
egin{align*}|intlimits_{gamma_R}e^{iz^p}{ m d}z|&=|int^frac{pi}{2p}_0 e^{i ho^pe^{ip heta}} ho e^{i heta}i{ m d} heta|\& ightarrow 0( ho ightarrow 0^+)end{align*}
于是可将$( ef{the1})$化为
egin{align*}int_0^{+infty} e^{ix^p}{ m d}x &=e^{ifrac{pi}{2p}}int_0^{+infty}e^{-x^p}\ &=e^{ifrac{pi}{2p}}int_0^{+infty}e^{-t}t^{frac{1}{p}-1}{ m d}t\ &=frac{1}{p}Gamma(frac{1}{p})e^{ifrac{pi}{2p}}end{align*}
故
egin{align}int_0^{+infty}cos x^p { m d}x &=frac{1}{p}Gamma(frac{1}{p})cos frac{pi}{2p}\ int_0^{+infty}sin x^p { m 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:
egin{align*} int_0^infty sin left( x^n
ight)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}
ight)}int_0^infty left(int_0^infty
u^{-frac{1}{n}}e^{-xu}du
ight) sin(x) dx\
&= frac{1}{n
Gamma left( 1-frac{1}{n}
ight)} int_0^infty u^{-frac{1}{n}}
left( int_0^infty e^{-xu}sin(x) dx
ight)du\
&= frac{1}{n Gamma left( 1-frac{1}{n}
ight)} int_0^infty frac{u^{-frac{1}{n}}}{1+u^2}du \
&=
frac{1}{n Gamma left( 1-frac{1}{n}
ight)}
int_0^{frac{pi}{2}} an^{-frac{1}{n}}( heta) d heta quad (u= an
heta) \
&= frac{1}{n Gamma left(
1-frac{1}{n}
ight)}int_0^{frac{pi}{2}}sin^{-frac{1}{n}}( heta)
cos^{frac{1}{n}}( heta) d heta \
&= frac{1}{2n Gamma left( 1-frac{1}{n}
ight)} mathrm{B} left( frac{1-n}{2},frac{1+n}{2}
ight) \
&=
frac{1}{2n Gamma left( 1-frac{1}{n}
ight)} Gamma left(
frac{n-1}{2n}
ight)Gamma left( frac{n+1} {2n}
ight) \
&= frac{sin left( frac{pi}{n}
ight)}{2ncos left( frac{pi}{2n}
ight)}Gamma left( frac{1}{n}
ight) \
&= frac{1}{n}sin left(frac{pi }{2n}
ight)Gamma left( frac{1}{n}
ight)end{align*}