武汉大学2011年数学分析试题解答

武汉大学2011年数学分析试题解答

1：计算题

（1）  解：原极限$\text{=}\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[n]{n}}{n}\cdot {{n}^{1-\alpha }}={{e}^{-1}}\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{1-\alpha }}=$$\left\{\begin{array}
+\infty, & \hbox{$0<\alpha<1$;} \\
e^{-1}, & \hbox{$\alpha=1$;} \\
 0, & \hbox{$\alpha>1$.}
\end{array}
\right.$
    (解释一下：$\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[n]{n}}{n}={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{i=1}^{n}{In\frac{i}{n}}}}={{e}^{\int\limits_{0}^{1}{Inxdx}}}={{e}^{-1}}$ （来源于数学分析上册第二章课后习题）

（2）  解：考虑等价无穷小：$\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{\frac{1}{2}{{x}^{2}}}=1$ ，则

$1-\cos \sqrt{\tan x-\sin x}=2{{\sin }^{2}}\frac{\sqrt{\tan x-\sin x}}{2}\sim \frac{1}{2}(\tan x-\sin x)=\frac{\sin x}{2\cos x}(1-\cos x)\sim \frac{1}{4}{{x}^{3}}$

另一方面：\[\sqrt[3]{1+{{x}^{3}}}-\sqrt[3]{1-{{x}^{3}}}=\frac{2{{x}^{3}}}{{{(\sqrt[3]{1+{{x}^{3}}})}^{2}}+\sqrt[3]{(1+{{x}^{3}})(1-{{x}^{3}})}+{{(\sqrt[3]{1-{{x}^{3}}})}^{2}}}\sim \frac{2{{x}^{3}}}{3}\]

从而原式   =$\frac{3}{8}$

（3）  法一：

解：原式\[=\int{\frac{1+\cos x}{\sqrt{1+\cos x}}dx=}\int{\frac{2{{\cos }^{2}}\frac{x}{2}}{\sqrt{1+\cos x}}dx=}\int{\frac{2{{\cos }^{2}}\frac{x}{2}(1-{{\sin }^{2}}\frac{x}{2})}{{{\cos }^{2}}\frac{x}{2}\sqrt{1+\cos x}}dx}\]      

      \[=\int{\frac{1+\cos x-\sin x\sin \frac{x}{2}\cos \frac{x}{2}}{{{\cos }^{2}}\frac{x}{2}\sqrt{1+\cos x}}dx=}\int{\sqrt{1+\cos x}{{\sec }^{2}}\frac{x}{2}+\frac{-\sin x}{\sqrt{1+\cos x}}\tan \frac{x}{2}dx}\]

      $=2\int{d(\sqrt{1+\cos x}\tan \frac{x}{2})=2}\sqrt{1+\cos x}\tan \frac{x}{2}+C$

法二：由于$\int{\sqrt{\text{1+}\cos x}dx}=\sqrt{2}\int{\left| \cos \frac{x}{2} \right|}dx$

考虑到$\int{\left| x \right|}dx=\frac{{{x}^{2}}}{2}sgn x+C$

于是$\int{\sqrt{\text{1+}\cos x}dx}=2\sqrt{2}\sin \frac{x}{2}sgn (\cos \frac{x}{2})+C$（$C$为常数）

法三：由于$\int{\sqrt{\text{1+}\cos x}dx}=\sqrt{2}\int{\left| \cos \frac{x}{2} \right|}dx$

$\overset{t=\frac{x}{2}}{\mathop{=}}\,2\sqrt{2}\int{\left| \cos t \right|}dt$

而

$\int {\left| {\cos t} \right|} dt=\left\{\begin{array}{ll}
\sin t + {c_k}, & \hbox{$- \frac{\pi }{2} + 2k\pi \le t \le \frac{\pi }{2} + 2k\pi$;} \\
- \sin t + {d_k}, & \hbox{$\frac{\pi }{2} + 2k\pi \le t \le \frac{{3\pi }}{2} + 2k\pi$.}
\end{array}
\right.$+C为连续函数，其中C为常数

于是

$\left\{\begin{array}{ll}
{c_k} + 1 = - 1 + {d_k} \\
{c_{k + 1}} - 1 = 1 + {d_k}
\end{array}
\right.$

，令${{c}_{0}}=0$

则${{c}_{k}}=4k,{{d}_{k}}=4k+2$

于是

$\int {\left| {\cos t} \right|} dt=\left\{\begin{array}{ll}
\sin t + 4k, & \hbox{$ - \frac{\pi }{2} + 2k\pi \le t \le \frac{\pi }{2} + 2k\pi$;} \\
- \sin t + 4k + 2, & \hbox{$\frac{\pi }{2} + 2k\pi \le t \le \frac{{3\pi }}{2} + 2k\pi$.}
\end{array}
\right.$+C

即

$\int {\sqrt {{\rm{1 + }}\cos x} dx}=\left\{\begin{array}{ll}
2\sqrt 2 (\sin \frac{x}{2} + 4k), & \hbox{$ - \pi + 4k\pi \le x \le \pi + 4k\pi$;} \\
2\sqrt 2 ( - \sin \frac{x}{2} + 4k + 2), & \hbox{$\pi + 4k\pi \le x \le 3\pi + 4k\pi$.}
\end{array}
\right.$+C，其中C为常数

（4）  解：\[F(x,y)=x\int_{\frac{y}{x}}^{xy}{zf(z)dz-y\int_{\frac{y}{x}}^{xy}{f(z)dz}}\]

则$F_{x}^{'}=\int_{\frac{y}{x}}^{xy}{zf(z)dz+x[xyf(xy)\cdot y-\frac{y}{x}f(\frac{y}{x})\cdot \frac{-y}{{{x}^{2}}}]-y[f(xy)\cdot y-f(\frac{y}{x})\cdot \frac{-y}{{{x}^{2}}}]}$

  $=\int_{\frac{y}{x}}^{xy}{zf(z)dz+({{x}^{2}}-1){{y}^{2}}f(xy)}$

  $F_{xx}^{''}=xyf(xy)\cdot y-\frac{y}{x}f(\frac{y}{x})\cdot \frac{-y}{{{x}^{2}}}+2x{{y}^{2}}f(xy)+({{x}^{2}}-1){{y}^{2}}f'(xy)\cdot y$

   $=3x{{y}^{2}}f(xy)+\frac{{{y}^{2}}}{{{x}^{3}}}f(\frac{y}{x})+({{x}^{2}}-1){{y}^{3}}f'(xy)$

（5）  解：原式$=\int\limits_{0}^{1}{dy\int\limits_{-1}^{{{y}^{2}}}{({{y}^{2}}-x)dx+}}\int\limits_{0}^{1}{dy\int\limits_{{{y}^{2}}}^{1}{(-{{y}^{2}}+x)dx}}=\frac{6}{5}$

2：说明，原版的试卷中的题目可能有点问题，原版试题如下：

已知$f(x),g(x)$在$[a,b]$上连续，在$(a,b)$上可微，且$g'(x)$在$(a,b)$上无零点，证明：$\exists \xi \in (a,b),st$ $\frac{f'(\xi )}{g'(\xi )}=\frac{f(b)-g(\xi )}{g(\xi )-g(a)}$

如果有思路的话，欢迎补充！

证明:作辅助函数$F\left( x \right)=f\left( x \right)g\left( x \right)-g\left( b \right)f\left( x \right)-f\left( a \right)g\left( x \right) $

虽然$F\left( x \right) $在$[a,b]$上连续，在$(a,b)$上可微, $F\left( a \right)=F\left( b \right)=-f\left( a \right)g\left( b \right) $

由罗尔中值定理,存在$\xi \in \left( a,b \right)$使得${F}'\left( \xi  \right)=0$

即${f}'\left( \xi  \right)g\left( \xi  \right)+f\left( \xi  \right){g}'\left( \xi  \right)-g\left( b \right){f}'\left( \xi  \right)-f\left( a \right){g}'\left( \xi  \right)=0$

整理$\left[ f\left( a \right)-f\left( \xi  \right) \right]{g}'\left( \xi  \right)-{f}'\left( \xi  \right)\left[ g\left( \xi  \right)-g\left( b \right) \right]=0$

即$\frac{f\left( a \right)-f\left( \xi  \right)}{g\left( \xi  \right)-g\left( b \right)}=\frac{{f}'\left( \xi  \right)}{{g}'\left( \xi  \right)} $,证得

3：（方法一）

证明：$\left| \frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}-b \right|=\left| \frac{{{a}_{1}}({{b}_{n}}-b)+{{a}_{2}}({{b}_{n-1}}-b)+\cdots +{{a}_{n}}({{b}_{1}}-b)}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}} \right|$ 

      $\le \frac{{{a}_{1}}\left| {{b}_{n}}-b \right|+\cdots +{{a}_{n-N}}\left| {{b}_{N+1}}-b \right|}{{{a}_{1}}+\cdots +{{a}_{n}}}+\frac{{{a}_{n-N+1}}\left| {{b}_{N}}-b \right|+\cdots +{{a}_{n}}\left| {{b}_{1}}-b \right|}{{{a}_{1}}+\cdots +{{a}_{n}}}$

      $\le \underset{N+1\le k\le n}{\mathop \max }\,\left| {{b}_{k}}-b \right|+\underset{N+1\le k\le n}{\mathop \max }\,\left| {{b}_{k}}-b \right|\cdot \frac{N}{n-N+1}=I_{1}^{n}+I_{2}^{n}$

从而对$\forall \varepsilon >0,$先取定$N$使得$I_{1}^{n}<\frac{\varepsilon }{2}$，后让$n$充分大即有$I_{2}^{n}<\frac{\varepsilon }{2}$，于是有结论成立。

（方法二）

证明：设${{t}_{nk}}=\frac{{{a}_{n-k+1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}},k=1,2,\cdots ,n,n=1,2,\cdots $

则${{t}_{nk}}>0$且$\sum\limits_{k=1}^{n}{{{t}_{nk}}}=1$

再由${{a}_{n}}\ge 0,{{a}_{n}}\le {{a}_{n-1}}\Rightarrow {{a}_{1}}+{{a}_{2}}+\cdots \text{+}{{a}_{n}}\ge {{a}_{1}}+{{a}_{2}}+\cdots \text{+}{{a}_{n-k+1}}\ge (n-k+1){{a}_{n-k+1}}$

于是$0\le {{t}_{nk}}\le \frac{{{a}_{n-k+1}}}{(n-k+1){{a}_{n-k+1}}}=\frac{1}{(n-k+1)}\to 0(n\to +\infty )$

由迫敛性知：$\underset{n\to +\infty }{\mathop{\lim }}\,{{t}_{nk}}=0$

再由$\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}=b$，可知

$\exists M>0,s.t$对任意的$n\in {{N}^{*}},\left| {{a}_{n}}-a \right|<M$

同时，对任意的$\varepsilon >0,\exists {{N}_{1}}\in {{N}^{*}}$，当$n>{{N}_{1}}$时，有$\left| {{a}_{n}}-a \right|<\frac{\varepsilon }{2}$

固定${{N}_{1}}$，由$\underset{n\to +\infty }{\mathop{\lim }}\,{{t}_{nk}}=0$可知：

$\exists {{N}_{2}}\in {{N}^{*}}$，当$n>{{N}_{2}}$时，有$\left| {{a}_{n}}-a \right|<\frac{\varepsilon }{2{{N}_{1}}M},k=1,2,\cdots ,{{N}_{1}}$

令$N=\max \{{{N}_{1}},{{N}_{2}}\}$，当$n>N$时，有

$\left| \frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}-b \right|=\left| \sum\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-b} \right|=\left| \sum\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-\sum\limits_{k=1}^{n}{{{t}_{nk}}}b} \right|$

$\le \sum\limits_{k=1}^{n}{{{t}_{nk}}}\left| {{b}_{k}}-b \right|<M({{t}_{n1}}+{{t}_{n2}}+\cdots +{{t}_{n{{N}_{1}}}})+\frac{\varepsilon }{2}({{t}_{n,{{N}_{1}}+1}}+\cdots +{{t}_{nn}})<\varepsilon $

即$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}=b$

（方法三，书上并无介绍此公式，不建议使用）：

直接利用$Stolz$进行计算即可

4：证明：${{a}_{k}}=\frac{1}{2\pi }\int\limits_{-\pi }^{\pi }{f(x)\cos kxdx=}\frac{1}{2k\pi }\int\limits_{-\pi }^{\pi }{f(x)d(sinkx)=}\frac{-1}{2k\pi }\int\limits_{-\pi }^{\pi }{f'(x)sinkxdx}$

         $=\frac{-1}{2{{k}^{2}}\pi }\int\limits_{-k\pi }^{k\pi }{f'(\frac{t}{k})sintdt}$

   则

${{a}_{2n}}=\frac{-1}{8{{n}^{2}}\pi }\int\limits_{-2n\pi }^{2n\pi }{f'(\frac{t}{2n})\operatorname{sint}dt=}\frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{[\int\limits_{-2m\pi }^{(2m+1)\pi }{f'(\frac{t}{2n})\operatorname{sint}dt+\int\limits_{(2m+1)\pi }^{2(m+1)\pi }{f'(\frac{t}{2n})\operatorname{sint}dt]=}}}$

\[\ge \frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{[\int\limits_{2m\pi }^{(2m+1)\pi }{f'(\frac{t}{2n})\sin tdt+}}\int\limits_{(2m+1)\pi }^{2(m+1)\pi }{\operatorname{sintdt}]}\] 

而

$\sin t\left\{\begin{array}{ll}
> 0, & \hbox{$ 2m\pi < x < (2m + 1)\pi$;} \\
< 0, & \hbox{$(2m + 1)\pi < x < 2(m + 1)\pi$.}
\end{array}
\right.$

 且$f$为凸函数

从而原式$\ge \frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{f'(\frac{2m+1}{2n}\pi )\int\limits_{2m\pi }^{2(m+1)\pi }{\sin tdt=0}}$

另一方面：

\[{{a}_{2n+1}}=\frac{-1}{2{{(2n+1)}^{2}}\pi }\int\limits_{-(2n+1)\pi }^{(2n+1)\pi }{f'(\frac{t}{2n+1})\sin tdt}\]

\[=\frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{[\int\limits_{(2m-1)\pi }^{2m\pi }{f'(\frac{t}{2n+1})\sin tdt+\int\limits_{2m\pi }^{(2m+1)\pi }{f'(\frac{t}{2n+1})\sin tdt}}}\]

$\le \frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{[\int\limits_{(2m+1)\pi }^{2m\pi }{f'(\frac{2m}{2n+1})\sin tdt}}+\int\limits_{2m\pi }^{(2m+1)\pi }{f'(\frac{2m}{2n+1})\sin tdt}]$

$==\frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{f'(\frac{2m}{2n+1}\pi )}\int\limits_{(2m-1)\pi }^{(2m+1)\pi }{\sin tdt=0}$

5：证明：

（1）  设${{f}_{k}}(x)=\sum\limits_{n=1}^{k}{{{u}_{n}}(x),}$

因为

 $\left| f(x)-f(y) \right|\le \left| f(x)-{{f}_{K}}(x) \right|+\left| {{f}_{K}}(x)-{{f}_{K}}(y) \right|+\left| {{f}_{K}}(y)-f(y) \right|$

 $\le 2\cdot \sup \left| \left| {{f}_{K}}(z)-f(z) \right| \right|+\left| {{f}_{K}}(x)-{{f}_{K}}(y) \right|\equiv I_{1}^{K}+I_{2}^{K}$

 从而对$\forall \varepsilon >0,$可先选定$K$使$I_{1}^{K}<\frac{\varepsilon }{2}$，后取$\delta >0$ 使\[\left| x-y \right|<\delta \Rightarrow I_{2}^{K}<\frac{\varepsilon }{2}\]

（2）  若条件改为逐点连续，则结论不成立。为此，仅需构造出${{f}_{k}}(x)\to f(x)$

其中每个${{f}_{k}}(x)$一致收敛，但$f(x)$不一致收敛，甚至不连续的例子很多，例如：

${f_k}(x)\left\{\begin{array}{ll}
- 1, & \hbox{$x \le \frac{{ - 1}}{k}$;} \\
kx, & \hbox{$\frac{{ - 1}}{k} < x < \frac{1}{k}$;} \\
1, & \hbox{$x \ge \frac{1}{k}$.}
\end{array}
\right.$

，恒有${{f}_{k}}(x)\to sgn (x)$

6：证明：

（1）$\int\limits_{c}^{d}{f(x,y)dy}$一致收敛

$\Leftrightarrow \exists F(x),\forall \varepsilon >0,\exists \delta >0,s.t.$
$\left\{\begin{array}{ll}
x \in [a,b] \\
\delta ' \in (0,\delta )
\end{array}
\right.$

其有$Cauchy$准则：

$\forall \varepsilon >0,\exists \delta >0,s.t.$
$\left\{\begin{array}{ll}
x \in [a,b] \\
\delta ',\delta '' \in (0,\delta )
\end{array}
\right.$
$\Rightarrow \left| \int\limits_{c+\delta '}^{c+\delta ''}{f(x,y)dy} \right|<\varepsilon $

（3）  仅需注意到$\left| \int\limits_{c+\delta '}^{c+\delta }{f(x,y)g(x,y)dy} \right|\le \underset{[a,b]\times [c,d]}{\mathop{\max }}\,\left| g(x,y) \right|\cdot \int\limits_{c+\delta '}^{c+\delta ''}{\left| f(x,y) \right|}dy$

 后利用$Cauchy$即可得证

7：（1）解：$g=f(\frac{x}{z})\mp zf(\frac{z}{x})+f(\frac{y}{z})\mp zf(\frac{z}{y})$

           ${{g}_{x}}=\frac{1}{z}f''(\frac{x}{z})\mp \frac{{{z}^{2}}}{{{x}^{2}}}f''(\frac{z}{x})$

           ${{g}_{xx}}=\frac{1}{{{z}^{2}}}f''(\frac{x}{z})\mp \frac{2{{z}^{2}}}{{{x}^{3}}}f'(\frac{z}{x})\mp \frac{{{z}^{3}}}{{{x}^{4}}}f''(\frac{z}{x})$

           ${{g}_{yy}}=\frac{1}{{{z}^{2}}}f''(\frac{y}{z})\mp \frac{{{z}^{2}}}{{{y}^{3}}}f'(\frac{z}{y})\mp \frac{{{z}^{3}}}{{{y}^{4}}}f''(\frac{z}{y})$

           ${{g}_{z}}=\frac{-1}{{{z}^{2}}}f'(\frac{x}{z})\mp f(\frac{z}{x})\mp \frac{z}{x}f'(\frac{z}{x})$

              $=\frac{-y}{{{z}^{2}}}f'(\frac{y}{z})\mp f(\frac{z}{y})\mp \frac{z}{y}f'(\frac{z}{y})$

           ${{g}_{zz}}=\frac{2x}{{{z}^{3}}}f'(\frac{x}{z})+\frac{{{x}^{2}}}{{{z}^{4}}}f''(\frac{x}{z})\mp \frac{1}{x}f'(\frac{z}{x})\mp \frac{1}{x}f'(\frac{z}{x})\mp \frac{z}{{{x}^{2}}}f''(\frac{z}{x})$

             $+\frac{2y}{{{z}^{3}}}f'(\frac{y}{z})+\frac{{{y}^{2}}}{{{z}^{4}}}f''(\frac{y}{z})\mp \frac{1}{y}f'(\frac{z}{y})\mp \frac{1}{y}f'(\frac{z}{y})\mp \frac{z}{{{y}^{2}}}f''(\frac{z}{y})$

  于是

            ${{x}^{2}}{{g}_{xx}}+{{y}^{2}}{{g}_{yy}}-{{z}^{2}}{{g}_{zz}}=-\frac{2x}{z}f(\frac{x}{z})-\frac{2y}{z}f(\frac{y}{z})$

（2）设$0<{{a}_{1}}<{{a}_{2}},0<{{b}_{1}}<{{b}_{2}},1<{{c}_{1}}<{{c}_{2}}.$则

  原式\[=-2\iiint_{\Omega }{[\frac{x}{z}f'(\frac{x}{z})+\frac{y}{z}f'(\frac{y}{z})dxdydz}\]

      \[=-2\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\int\limits_{{{b}_{1}}}^{{{b}_{2}}}{dv\int\limits_{{{c}_{1}}}^{{{c}_{2}}}{[\frac{1}{u}f'(\frac{1}{u})+\frac{1}{v}f'(\frac{1}{v})]\cdot \frac{2}{3}{{u}^{-2}}{{v}^{-2}}dw}}}\]

  （其中令$u=\frac{z}{x},v=\frac{z}{y},w={{z}^{3}}$ ）

      $\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\int\limits_{{{b}_{1}}}^{{{b}_{2}}}{[\frac{1}{{{u}^{2}}v}f'(\frac{1}{u})+\frac{1}{u{{v}^{2}}}f'(\frac{1}{v})}}]dv$

     \[=\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{\{\frac{1}{{{u}^{2}}}f'(\frac{1}{u})In\frac{{{b}_{2}}}{{{b}_{1}}}-\frac{1}{u}[f(\frac{1}{{{b}_{2}}})-f(\frac{1}{{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}du}\] 

     \[=\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\left\{ -In\frac{{{b}_{2}}}{{{b}_{1}}}[ \right.f(\frac{1}{{{a}_{2}}})-f(\frac{1}{{{a}_{1}}})]-[f(\frac{1}{{{b}_{2}}})-f(\frac{1}{{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}\]

     \[=\frac{2({{c}_{2}}-{{c}_{1}})}{3}\left\{ In\frac{{{b}_{2}}}{{{b}_{1}}}[ \right.f(\frac{1}{{{a}_{2}}})+f(\frac{1}{{{a}_{1}}})]-[f(\frac{1}{{{b}_{2}}})-f(\frac{1}{{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}\]

  8：证明：\[\iint_{D}{(x\frac{\partial u}{\partial x}}+y\frac{\partial u}{\partial y})dxdy=\int\limits_{0}^{1}{dr}\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{(}x{{u}_{x}}+y{{u}_{y}})dS\]

           \[=\int\limits_{0}^{1}{r}\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{\frac{\partial u}{\partial n}}dS=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\nabla udxdy}\]

           \[=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\cos (\pi ({{x}^{2}}+{{y}^{2}}))dxdy}\]

           \[=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\cos (\pi {{s}^{2}})\cdot 2\pi sdS}\]

           \[=\int\limits_{0}^{1}{r\sin (\pi r)dr=\frac{1}{\pi }}\]