一些数学分析考研题（2024年）的解析

证明 \[\because \lim_{x\to+\infty }f(x)存在.\]\[\therefore \forall \varepsilon > 0，\exists X,x',x''> X,(x',x''\in [X+1,+\infty )),对于\delta > 0,0< ｜x'-x''｜< \delta，\]\[ s.t.|f(x')-f(x'')|< \varepsilon . 一致连续.\]又因为\[f(x)\in C[a,+\infty ).所以对于上面的\varepsilon ，\forall x',x''\in [a,X+1],对于\delta > 0,0< ｜x'-x''｜< \delta，\]\[ s.t.|f(x')-f(x'')|< \varepsilon . 一致连续.\]从而有\[f(x)在[a,+\infty )上一致连续.\]
        [注]此结论。相当有用。可以总结为：在某开区间内I连续的函数，且在I的左、右或左右两个端点极限存在，那么。函数在该区间I内一致连续。

证明 由\[f(x)=x^3e^{-x^2}.\]得\[f'(x)=3x^2e^{-x^2}-2x^4e^{-x^2}=0.x=\pm \sqrt{\frac{3}{2}}.\]\[f''(x)=6xe^{-x^2}-6x^3e^{-x^2}-8x^3e^{-x^2}+4x^5e^{-x^2}.f''(\frac{3}{2})< 0,f''(-\frac{3}{2})>  0.\]得到\[f(-\frac{3}{2})\le f(x)\le f(\frac{3}{2}).\]计算函数极大极小值 \[f(\frac{3}{2})=\frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},\]\[f(-\frac{3}{2})=-\frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},\]于是\[|f(x)|\le \frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},函数f(x)有界.\]

证明 由于\[A> 0,x\in [\delta ,+\infty ).|\int_{0}^{A}\sin (xy)dy|\le |\frac{1-\cos (\delta A)}{x}|\le \frac{2}{\delta }.\]\[\frac{1}{y}\downarrow ,\rightarrow 0.\]由Dirichlet判别法，\[\int_{0}^{+\infty }\frac{\sin (xy)}{y}dy在[\delta ,+\infty )上一致收敛.\]

      \[对M> 0,总能找到A_1,A_2> M.比如A_1=\frac{\pi }{3}M,A_2=\frac{2\pi }{3}M.x=\frac{1}{M}\in (0,+\infty ).\]由柯西收敛定理\[|\int_{A_1}^{A_2}\frac{\sin (xy)}{y}dy|\geqslant \frac{1}{2}|\int_{\frac{\pi }{3}M}^{\frac{2\pi }{3}M}\frac{1}{y}dy|=\frac{\ln2}{2}=\varepsilon _0> 0.\]因此\[\int_{0}^{+\infty }\frac{\sin (xy)}{y}dy在[0 ,+\infty )上非一致收敛.\]

证明  由于\[\mathrm{E}有界，\forall x_n\in \mathrm{E},a=M_1\le x_n\le M_2=b.即x_n\in [a,b]=[a_0,b_0].\]\[
中分[a,b]，得到两个区间，则必有一个部分包含\mathrm{E}的无穷多个点x_n,设这个区间为[a_1,b_1].\]\[如此下去，得到[a_2,b_2]，\cdots ，[a_n,b_n].\]由取法，显然有\[[a,b]=[a_0,b_0]\supset [a_1,b_1]\supset [a_2,b_2]\supset \cdots \supset [a_n,b_n].\]从而有\[\lim_{n\to\infty }(b_n-a_n)=0.\]\[\exists x\in \mathrm{E},s.t.\lim_{n\to\infty }b_n=\lim_{n\to\infty }a_n=x.\]
同时也可知，\[至少存一个无穷子列\{a_{n_k}\}(或\{b_{n_k}\})趋于x,x即为聚点.\]

证明 由法平面与积分元的关系，有\[ \cos \alpha dS=dydz,\cos \beta dS=dzdx,\cos \gamma dS=dxdy.\]\[\begin{align*}\therefore I
&=\frac{1}{3}\iint\limits_{S} (x\cos \alpha+y\cos \beta +z\cos \gamma )dS\\
&=\frac{1}{3}\iint\limits_{S} xdydz+ydzdx+zdxdy\\
&=\frac{1}{3}\iiint\limits_{\Omega }(1+1+1)dV \\
&=\varDelta V.
\end{align*}\]

解\[\begin{align*}(x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}
&=(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2}\frac{1}{x^2}+\frac{1}{6}\frac{1}{x^3}+o(\frac{1}{x^3})) \\
&=(x^3-x^2+\frac{x}{2})+(x^2-x+\frac{1}{2})+\frac{1}{2}(x-1+\frac{1}{2x})+\frac{1}{6}(1-\frac{1}{x}+\frac{1}{2x^2}) \\
&=x^3+\frac{1}{6}+\frac{1}{12x}+\frac{1}{12x^2}.
\end{align*}\]\[\sqrt{1+x^6}=x^3\sqrt{1+x^{-6}}=x^3(1+\frac{1}{2}x^{-6})=x^3+\frac{1}{2}x^{-3}.\]\[\lim_{x\to\infty }[(x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}-\sqrt{1+x^6}]=\lim_{x\to\infty }[x^3+\frac{1}{6}+\frac{1}{12x}+\frac{1}{12x^2}-(x^3+\frac{1}{2}x^{-3})]=\frac{1}{6}.\]

解\[偶延拓，b_n=0.a_0=\frac{2}{\pi }\int_{0}^{\pi }\sin xdx=\frac{4}{\pi }.\]\[\begin{align*}a_n
&=\frac{2}{\pi }\int_{0}^{\pi }\sin x\cos nxdx \\
&=\frac{2}{\pi }\int_{0}^{\pi }[\sin (n+1)x+\sin (1- n)x]dx \\
&=\frac{1}{\pi } [\frac{1}{1+n}(1-\cos (n+1)\pi )+\frac{1}{1-n}(1-\cos (1-n)\pi )].
\end{align*}\]注意到\[n=2k+1,a_n=0.\]\[n=2k,a_n=\frac{4}{\pi }\frac{1}{1-4k^2}.\]所以\[f(x)=\sin x\backsim \frac{a_0}{2}+\displaystyle\sum_{n=1}^{\infty }a_n\cos nx=\frac{2}{\pi }+\frac{4}{\pi }\displaystyle\sum_{k=1}^{\infty }\frac{1}{1-4k^2}\cos 2kx.\]

解 变量变换，令\[u=x+y,v=y.0\le v\le u,0\le u\le 1.\]\[\frac{\partial(u,v)}{\partial (x,y)}=\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}=\begin{vmatrix}
1 & 1\\
0 & 1
\end{vmatrix}=1.\]\[I=\iint\limits_{D}e^{\frac{y}{x+y}}dxdy=\int_{0}^{1}du\int_{0}^{u}e^{\frac{v}{u}}dv=\int_{0}^{1}udu\int_{0}^{u}e^{\frac{v}{u}}d(\frac{v}{u})=\int_{0}^{1}u[e-1]du=\frac{1}{2}(e-1).\]

解\[\begin{align*}I
&=\oint_{L}\frac{(x+y)dx-(x-y)dy}{x^2+y^2} \\
&=\frac{1}{a^2}\oint_{L}(x+y)dx-(x-y)dy \\
&= \frac{1}{a^2}\iint_{D}(-1-1)dxdy,(格林公式)\\
&=-2\pi .
\end{align*}\]

解\[\begin{align*}I
&=\iint\limits_{S}xdydz+ydzdx+zdxdy \\
&=\iiint\limits_{\Omega }(1+1+1)dV \\
&=3\cdotp \frac{4}{3}\pi a^3 \\
&=4 \pi a^3.
\end{align*}\]