来自浙大2013考研数学分析的一个问题

陈洪葛 posted @ Jun 22, 2013 12:05:58 PM in 数学分析 , 955 阅读

证明
\[ \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}}=0 \]



\[ I=\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}=\frac{1}{n}\cdot\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{y^{\frac{1}{n}-1}\sin{y}dy}\qquad (y=t^{n}) \]

\[ \Gamma\left(1-\frac{1}{n}\right)=\int_{0}^{\infty}{u^{-\frac{1}{n}}e^{-u}du}=\int_{0}^{\infty}{y^{1-\frac{1}{n}}\cdot x^{-\frac{1}{n}}e^{-xy}dx} \]
\begin{align*}
 I&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{\left(\int_{0}^{+\infty}{x^{-\frac{1}{n}}e^{-xy}dx}  \right)\cdot\sin{y}dy}\\
&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{+\infty}{x^{-\frac{1}{n}}\cdot\left(\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{e^{-xy}\sin{y}dy}  \right)dx}\\
&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}-\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{\infty}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\\
&=I_{1}-I_{2}
\end{align*}
而我们知道
\[ \int_{0}^{+\infty}{\frac{t^{a-1}}{1+t}dt}=\frac{\pi}{\sin{\pi a}}\qquad  (\text{Euler}) \]
\[ \Rightarrow \int_{0}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}=\frac{\pi}{2\cos{\frac{1}{2n}}}\]
\[ \Rightarrow \lim_{n\to\infty}{I_{1}}=0 \]
\[ I_{2}=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot\left(\int_{0}^{1}+\int_{1}^{+\infty}\right)=S_{1}+S_{2}\]
\begin{align*}
|S_{1}|&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\left|\int_{0}^{1}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx} \right|\\
&\leq\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{1}{\frac{[\left(\frac{\pi}{2}\right)^{n}+1]x^{-\frac{1}{n}}}{e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\\
&= \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \frac{\left(\frac{\pi}{2}\right)^{n}+1}{\left(\frac{\pi}{2}\right)^{n-1}}\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{z^{-\frac{1}{n}}e^{-z}dz}\\
&\leq \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \frac{\left(\frac{\pi}{2}\right)^{n}+1}{\left(\frac{\pi}{2}\right)^{n-1}}\int_{0}^{+\infty}{z^{-\frac{1}{n}}e^{-z}dz}\\
&=\frac{1}{n}\cdot\left[\left( \frac{\pi}{2}\right)+\left( \frac{\pi}{2}\right)^{1-n}  \right]\to 0
\end{align*}
\begin{align*}
 |S_{2}|&= \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\left|\int_{1}^{\infty}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\right|\\
&\leq \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \int_{1}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}\\
&\leq \frac{\pi}{2n\Gamma\left(1-\frac{1}{n}\right)}\to 0
\end{align*}

\[ \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}}=0 \]

——————————————————————————————————————————
顺便计算
\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*}

——————————————————————————————————————————————

显然,上面的证明弄复杂了,太非主流了。下面给出个主流的证明

对$\forall \varepsilon>0$,存在$ 0<a<\frac{\varepsilon}{4} $,注意到
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{\sin{t^n}dt}&= \int_{0}^{1-a}{\sin{t^n}dt}+ \int_{1-a}^{1+a}{\sin{t^n}dt}+ \int_{1+a}^{\frac{\pi}{2}}{\sin{t^n}dt}\\
&=I_{1}+I_{2}+I_{3}
\end{align*}
由$ |\sin{x}|\leq 1 $知
\[ |I_{2}|<\frac{\varepsilon}{2} \]
对$I_{1}$,有
\[ |I_{1}|\leq \int_{0}^{1-a}{|\sin{t^n}|dt}\leq \int_{0}^{1-a}{t^{n}dt}\leq \frac{(1-a)^{n+1}}{n+1} \]
显然可以找到一个$N_{1}>0$使得$n>N_{1}$时有
\[ |I_{1}|\leq \frac{\varepsilon}{4} \]
而对$I_{3}$
\begin{align*}
 I_{3}&=\int_{1+a}^{\frac{\pi}{2}}{\sin{t^n}dt}\\
&= \int_{1+a}^{\frac{\pi}{2}}{\frac{d(-\cos{t^n})}{nt^{n-1}}}\\
&=\frac{-\cos t^n}{nt^{n-1}}\bigg|_{1+a}^{\frac{\pi}{2}}+\frac{1-n}{n}\cdot \int_{1+a}^{\frac{\pi}{2}}{\frac{\cos{t^{n}}}{t^{n}}dt}\\
&=\frac{\cos{(1+a)^n}-\cos{\left(\frac{\pi}{2}\right)^n}}{n(1+a)^{n-1}}+\frac{1-n}{n}\cdot \int_{1+a}^{\frac{\pi}{2}}{\frac{\cos{t^{n}}}{t^{n}}dt}
\end{align*}
显然存在$N_{2}$使得当$n>N_{2}$时有
\[ |I_{3}|<\frac{\varepsilon}{4} \]
这样,取$N=\max\{N_{1},N_{2}\} $,当$n>N$时,就有
\[ |I|<\varepsilon \]
\[ \Rightarrow \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^n}}dt}=0 \]


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