数学分析习题练习【谢绝回贴】

解\[令r=\sqrt{a^2+b^2+c^2}.进行正交变换(x,y,z)\rightarrow (u,v,w).\]\[u=\frac{1}{r}(ax+by+cz).\Sigma :x^2+y^2+z^2=1,\rightarrow \Sigma ':u^2+v^2+w^2=1.\]\[v^2+w^2=1-u^2,v=\sqrt{1-u^2}\cos \theta ,w=\sqrt{1-u^2}\sin \theta.\]\[-1\le u\le 1,0\le \theta \le 2\pi.\]\[E=\frac{1}{1-u^2},F=0,G=1-u^2.\]\[dS=\sqrt{EG-F}dud\theta =dud\theta.\]\[\begin{align*}\therefore \iint\limits_{\Sigma }f(ax+by+cz)dS
&=\iint\limits_{\Sigma }f(ru)dS \\
&=\int_{0}^{2\pi} d\theta \int_{-1}^{1}f(\sqrt{a^2+b^2+c^2}u)du\\
&=2\pi \int_{-1}^{1}f(\sqrt{a^2+b^2+c^2}u)du.
\end{align*}\]

解 注意到\[\int_{a}^{b}e^{-xy}dy=\frac{e^{-ax}-e^{-bx}}{x}.\]\[\begin{align*}I(a,b)
&=\int_{0}^{+\infty }\frac{e^{-ax}-e^{-bx}}{x}\cos mx dx \\
&=\int_{0}^{+\infty }\cos mx dx\int_{a}^{b}e^{-xy}dy  \\
&=\int_{a}^{b}dy\int_{0}^{+\infty }e^{-xy}\cos mx dx,交换积分顺序 \\
&=\int_{a}^{b}\frac{y}{y^2+m^2}dy \\
&=\frac{1}{2} \ln(y^2+m^2).
\end{align*}\]

证明\[首先添加y=y_0直线段AB与上半园组成一个闭合区域\Sigma .\]\[(x_0,y_0)在AB上.用挖洞法，用半径为\varepsilon 的圆，于是利用格林公式，并由已知条件，得\]\[I(r)=[\oint_{\Sigma -C_r}+\oint_{C_r}-\int\limits_{AB}](Pdx+Qdy)=0.\]\[\oint_{\Sigma -C_r}Pdx+Qdy=0.(已知条件符合线积分与路径无关)\]\[\oint_{C_r}Pdx+Qdy=\iint_{\Sigma_r}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})|_{(x',y')}\cdotp \frac{\pi\varepsilon ^2}{2}.(中值定理)\]\[\int\limits_{AB}(Pdx+Qdy)=\int\limits_{AB}Pdx=P(x'',y_0)\cdotp 2\varepsilon .(中值定理)\]从而有\[(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})|_{(x',y')}\cdotp \frac{\pi\varepsilon }{2}=2P(x'',y_0).\]\[令\varepsilon \rightarrow 0^+，P(x'',y_0)=P(x_0,y_0).\Rightarrow P(x_0,y_0)=0.\]再者由\[P(x'',y_0)=0，所以(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})=\frac{\partial Q}{\partial x}，\]\[令\varepsilon \rightarrow 0^+，\frac{\partial Q(x_0,y_0)}{\partial x}=0.\]\[而由（x_0,y_0）的任意性，所以恒有P(x,y)=0，\frac{\partial Q(x,y)}{\partial x}=0.\]

证明 \[由黎曼引理\lim_{n\to\infty }\int_{a}^{b}f(x)g(nx)dx=\frac{1}{T}\int_{0}^{T}g(x)dx\int_{a}^{b}f(x)dx.\]\[其中g(x)是以 T为周期的函数.\]有\[\lim_{n\to\infty }\int_{0}^{a}f(x)\sin (nx)dx=\frac{1}{2\pi}\int_{0}^{2\pi}\sin xdx\int_{0}^{a}f(x)dx,\]\[\because f(x)单调，\therefore |\int_{0}^{a}f(x)dx|< M，(M> 0)\]因此\[\lim_{n\to\infty }\int_{0}^{a}f(x)\sin (nx)dx=\frac{1}{2\pi}\int_{0}^{2\pi}\sin xdx\int_{0}^{a}f(x)dx=0.\]同样的，当\[\lim_{n\to+\infty }f(x)=0,则\int_{0}^{+\infty }f(x)dx< \infty .\]\[\therefore \lim_{n\to\infty }\int_{0}^{+\infty }f(x)\sin (nx)dx=\frac{1}{2\pi}\int_{0}^{2\pi}\sin xdx\int_{0}^{+\infty }f(x)dx=0.\]

        基本与上题同。

2024北京师范大学数学分析


解  显然\[a_n> 0.a_{n+1}=\ln(1+a_n)< a_n.\{a_n\}收敛.\]令\[\lim_{n\to\infty }a_n=a.\Rightarrow a=0.\]\[\begin{align*}\lim_{n\to\infty }na_n
&=\lim_{n\to\infty }\frac{n+1-n}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}} \\
&=\lim_{n\to\infty }\frac{a_na_{n+1}}{a_n-a_{n+1}} \\
&=\lim_{n\to\infty }\frac{a_n\ln(1+a_n)}{a_n-\ln(1+a_n)} \\
&==\lim_{n\to\infty }\frac{a_n(a_n-\frac{1}{2}a^2_n)}{\frac{1}{2}a^2_n}\\
&=2.
\end{align*}\]

2024北京师范大学数学分析


证明    由已知\[\forall \varepsilon > 0，x',x''\in [a,+\infty ),当x'，x''> X,\exists \delta > 0,\]\[0< |x'-x''|< \delta=\frac{\varepsilon }{b} ,s.t.\]\[|f(x')-f(x'')|=|y'-y''|=b|x'-x''|< \varepsilon.\]          命题成立。