一些数学分析考研题（2024年）的解析

证明   \[\because \varphi (y)=\int_{a}^{+\infty }f(x,y)dx在y\in [c,d]上一致收敛，\]\[\therefore \forall \varepsilon > 0,\exists A> 0,|\int_{A}^{+\infty }f(x,y)dx|< \frac{\varepsilon }{3}.\]\[对于y,y_0\in [a,+\infty),\delta > 0,0< |y-y_0|< \delta ,s.t.\]\[|\int_{a}^{A}f(x,y)dx-\int_{a}^{A}f(x,y_0)dx|\le \int_{a}^{A}|f(x,y)dx-f(x,y_0)|dx< \frac{\varepsilon }{3}.\]因此\[\begin{align*}|\varphi (y)-\varphi (y_0)|
&=|\int_{a}^{A}f(x,y)dx+\int_{A}^{+\infty }f(x,y)dx-\int_{a}^{A}f(x,y_0)dx-\int_{A}^{+\infty }f(x,y_0)dx| \\
&\le |\int_{a}^{A}f(x,y)dx-\int_{a}^{A}f(x,y_0)dx|+ |\int_{A}^{+\infty }f(x,y)dx|+|\int_{A}^{+\infty }f(x,y_0)dx| \\
&<\frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3} \\
&=\varepsilon .
\end{align*}\]            而由$y_0$的任意性，可知$\varphi (y)\in C[c,d] .$


         

解  
         添加直线段$A(-2,0)B(2,0).$使之与BA弧组成闭合区域。原点$（0，0）$在闭合区域边界上,为奇点。作半经为$\varepsilon $的小圆，挖去小圆：$x^2+y^2=\varepsilon ^2.$\[\begin{align*}I
&=\int_{L}\frac{ydx-xdy}{x^2+y^2} \\
&=\oint_{L_{BA+AB}}\frac{ydx-xdy}{x^2+y^2}-\int_{BA}\frac{ydx-xdy}{x^2+y^2}\\
&=\oint_{L_{BA+AB-\frac{1}{2}C_\varepsilon }}\frac{ydx-xdy}{x^2+y^2}-\oint_{L_{\frac{1}{2}C_\varepsilon }}\frac{ydx-xdy}{x^2+y^2}-0 \\
&=0-\frac{1}{\varepsilon^2} \oint_{L_{C_\varepsilon }}ydx-xdy-0 \\
&=-\pi .
\end{align*}\]

解 \[I(a)=\int_{0}^{+\infty }\frac{e^{-x}\sin ax}{x}dx,\]\[\begin{align*}I'(a)
&=\int_{0}^{+\infty }\frac{xe^{-x}\cos  ax}{x}dx \\
&=-e^{-x}\cos  ax|_0^{+\infty }-a\int_{0}^{+\infty }e^{-x}\sin axdx \\
&=1+ae^{-x}\sin ax|_0^{+\infty }-a^2\int_{0}^{+\infty }e^{-x}\cos axdx \\
&=1-a^2\int_{0}^{+\infty }e^{-x}\cos axdx.
\end{align*}\]所以\[I'(a)=\int_{0}^{+\infty }\frac{xe^{-x}\cos  ax}{x}dx=\frac{1}{1+a^2}.\]\[I(a)=\int_{0}^{a}\frac{1}{1+a^2}da+I(0)=\arctan \alpha .\]

证明  \[p> 1,|\int_{1}^{+\infty }\frac{\sin x}{x^p}\cos \frac{1}{x}dx|\le \int_{1}^{+\infty }\frac{1}{x^p}dx< \infty .\]广义积分绝对收敛。而\[0< p\le 1,\begin{align*}|\int_{1}^{+\infty }\frac{\sin x}{x^p}\cos \frac{1}{x}dx|
&=|\int_{1}^{+\infty }\frac{\sin( x+\frac{1}{x})+\sin( x-\frac{1}{x})}{x^p}dx| \\
&\geqslant |\int_{1}^{+\infty }\frac{\sin^2( x+\frac{1}{x})+\sin^2( x-\frac{1}{x})}{x^p}dx| \\
&\geqslant |\int_{1}^{+\infty }\frac{2}{x^p}dx| \\
&=\infty .
\end{align*}\]广义积分非绝对收敛。但此时\[\because \int_{1}^{+\infty }\sin x\le M.\frac{1}{x^p}\downarrow \rightarrow 0,\]\[\therefore \int_{1}^{+\infty }\frac{\sin x}{x^p}\cos \frac{1}{x}dx\le \int_{1}^{+\infty }\frac{\sin x}{x^p}dx<  \infty .\]说明广义积分条件收敛。

解      可以直接计算。（略）

解  积分平面$\Sigma$ 的单位微量：$\vec{n}=\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\}.$由$Stoks$定理\[\begin{align*}I
&=\oint_{C}(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz \\
&=\iint\limits_{\Sigma }\begin{vmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\
y^2-z^2 &z^2-x^2  & x^2-y^2
\end{vmatrix}dS \\
&=-\frac{4}{\sqrt{3}}\iint\limits_{\Sigma }(x+y+z)dS\\
&=-\frac{4}{\sqrt{3}}\iint\limits_{\Sigma }dS\\
&=-\frac{4}{\sqrt{3}}\pi .
\end{align*}\]

证明 \[\begin{align*}d_n&=\int_{0}^{1}f(x)dx-\frac{1}{n}\displaystyle\sum_{k=1}^{n}f(\frac{k-1}{n})\\&=\displaystyle\sum_{k=1}^{n}\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(x)dx-\frac{1}{n}[\displaystyle\sum_{k=1}^{n}f(\frac{k}{n})+f(0)-f(1)]\\&=\displaystyle\sum_{k=1}^{n}\int_{\frac{k-1}{n}}^{\frac{k}{n}}[f(x)-f(\frac{k}{n})]dx+\frac{1}{n}[f(1)-f(0)]\\&=\displaystyle\sum_{k=1}^{n}\int_{\frac{k-1}{n}}^{\frac{k}{n}}f'(\xi _k)(x-\frac{k}{n})dx+\frac{1}{n}[f(1)-f(0)],(被积函数连续.\xi _k\in (\frac{k}{n},x))\\&=\displaystyle\sum_{k=1}^{n}f'(\xi _k)\int_{\frac{k-1}{n}}^{\frac{k}{n}}(x-\frac{k}{n})dx+\frac{1}{n}[f(1)-f(0)],(x\to\frac{k}{n})\\&=\displaystyle\sum_{k=1}^{n}f'(\xi _k)[\frac{2k-1}{2n^2}-\frac{k}{n^2}]+\frac{1}{n}[f(1)-f(0)]\\&=-\frac{1}{2n^2}\displaystyle\sum_{k=1}^{n}f'(\xi _k)+\frac{1}{n}[f(1)-f(0)]\\&=-\frac{1}{2n}\int_{0}^{1}f'(\xi )dx+\frac{1}{n}[f(1)-f(0)]\\&=\frac{1}{2n}[f(1)-f(0)].
\end{align*}\]\[\therefore \lim_{n\to\infty }nd_n=\frac{1}{2}[f(1)-f(0)].\]

东北大学2024
解\[\lim_{x\to0}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}=\lim_{x\to0}\frac{x-x^2+o(x^2)-x+\frac{1}{2}x^2+o(x^2)}{x^2}=-\frac{1}{2}.\]

东北大学2024
解 \[\begin{align*}\int_{0}^{1}x^2f(x)dx
&=\int_{0}^{1}x^2dx\int_{1}^{x}e^{-y^2}dy \\
&=\int_{0}^{1}e^{-y^2}dy\int_{0}^{y}x^2dx\\
&=\frac{1}{3}\int_{0}^{1}y^3e^{-y^2}dy \\
&=\frac{1}{6}\int_{0}^{1}y^2e^{-y^2}dy^2 \\
&=\frac{1}{6}\int_{0}^{1}te^{-t}dt,(y^2=t)\\
&=\frac{1}{6}[-te^{-t}|_0^1+\int_{0}^{1}e^{-t}dt]\\
&=\frac{1}{6}[-e^{-1}-e^{-1}+1]\\
&=\frac{1}{6}-\frac{1}{3}e^{-1}.
\end{align*}\]

东北大学2024


解\[\begin{align*}\displaystyle\sum_{n=0}^{\infty }\frac{(-1)^n(n^2-n+1)}{2^n}
&=\displaystyle\sum_{n=0}^{\infty }\frac{(-1)^nn(n-1)}{2^n}+\displaystyle\sum_{n=0}^{\infty }\frac{(-1)^n}{2^n} \\
&=x^2\displaystyle\sum_{n=0}^{\infty }(-1)^nn(n-1)x^{n-2}+\displaystyle\sum_{n=0}^{\infty }(-1)^nx^n,(x=\frac{1}{2})\\
&=x^2(\displaystyle\sum_{n=0}^{\infty }(-1)^nx^n)''+\displaystyle\sum_{n=0}^{\infty }(-1)^nx^n\\
&=x^2(\frac{1}{1+x})'' +\frac{1}{1+x}\\
&=\frac{2x^2}{(1+x)^3}+\frac{1}{1+x}\\
&=\frac{3x^2+2x+1}{(1+x)^3}\\
&=\frac{22}{27},(x=\frac{1}{2}代入)
\end{align*}\]