用户: Ruiqi chen/CoursesNotes/数学分析 I (实验班)/Oct 28 2024

1正文

定理 1.1 (微积分基本定理). 假定 $f$ 在 $[a,b]$ 可积, 定义$$F(x)={\int }_a^{x}f(t)dt$$(1)则 $F(x)$ 连续, 在 $f(x)$ 的连续点 $x_0$ 处可导, 且 $F^{\prime }(x)=f(x)$.

证明. 有 $\left|F(x)-F(x_0)\right|=\left|{\int }_{x_0}^x{F}^{\prime }(t)dt\right|\le \left|x-x_0\right|\left|\max \left\{f(x)\right\}\right|$, 可导性来自于: $$\begin{array}{rl}\frac{F(x)-F(x_0)}{x-x_0}& =\frac{1}{x-x_0}{\int }_{x_0}^x{F}^{\prime }(t)dt\\ \left|\frac{F(x)-F(x_0)}{x-x_0}-f(x_0)\right|& =\left|\frac{1}{x-x_0}{\int }_{x_0}^{x}f(t)-f(x_0)dt\right|\\ & \le \max \left\{f(t)-f(x)\right\}\to 0\end{array}$$

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定理 1.2 (Newton–Leibniz 公式). $F$ 在 $[a,b]$ 上可导, 导函数 $F^{\prime }$ Riemann 可积, 则: $${\int }_a^b{F}^{\prime }(x)dx=F(b)-F(a)$$(2)

证明. 有展开: $$F(b)-F(a)=\sum _{i=1}^n{F}^{\prime }({\xi }_i)(x_{i+1}-x_i)\to {\int }_a^b{F}^{\prime }(x)dx$$(3)

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定理 1.3 (分部积分). $${\int }_a^{b}F(x)G^{\prime }(x)dx=F(b)G(b)-F(a)G(a)-{\int }_a^b{F}^{\prime }(x)G(x)dx$$(4)

命题 1.4 (积分余项的 Taylor 公式). $f$ 在 $[a,b]$ 上 $m+1$ 次连续可微, 则$$f(b)=\sum _{k=0}^m\frac{f^{(k)}(a)}{k!}{(b-a)}^k+{\int }_a^b\frac{f^{(m+1)}(\xi )}{m!}{(b-x)}^{m}dx$$(5)

证明. 分部积分.

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命题 1.5. 有: $$\begin{array}{rl}\int x^{\alpha }dx& =\frac{x^{\alpha +1}}{\alpha +1}+C\\ \int \frac{1}{x}dx& =\ln \left|x\right|+C\\ \int \ln xdx& =x\ln x-x+C\\ \int \sin xdx& =-\cos x+C\\ \int \cos xdx& =\sin x+C\\ \int e^{x}dx& =e^x+C\\ \int \frac{1}{x^2+a^2}dx& =\frac{1}{a}\arctan \frac{x}{a}+C\\ \int \frac{1}{\sqrt{a^2-x^2}}dx& =\arcsin \frac{x}{a}+C\\ \int \tan axdx& =-\frac{1}{a}\ln \left|\cos ax\right|+C\\ \int \frac{1}{\sin x}dx& =\ln \left|\tan \frac{x}{2}\right|+C\\ \int \frac{1}{{\cos }^{2}ax}dx& =\frac{1}{a}\tan ax+C\end{array}$$

定理 1.6. 如果$$\int f(y)dy=F(y)+C$$(6)则有$$\int f(g(x))u^{\prime }(x)dx=F(u(x))+C$$(7)若 $x=u^{\prime }(t)\ne 0$, 则有: $$\int f(u(t))u^{\prime }(t)dt=G(t)+C⟹\int f(x)dx=G(u^{-1}(t))+C$$(8)

例 1.7 (Wallis 积分). 有: $$\underset{n\to +\infty }{\lim }\sqrt{\frac{2n}{\pi }}{I}_n=1,I_n:={\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx$$(9)

证明. 有: $$\begin{array}{rl}{I}_{n+2}-I_n& ={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x({\sin }^{2}x-1)dx\\ & =-{\int }_0^{\frac{\pi }{2}}{\sin }^{n}x{\cos }^{2}xdx\\ & =-{\int }_0^{\frac{\pi }{2}}{\sin }^{n}x\cos xd\sin x\\ & =-\frac{1}{n+1}{\int }_0^{\frac{\pi }{2}}\cos xd{\sin }^{n+1}x\\ & =\frac{1}{n+1}{\int }_0^{\frac{\pi }{2}}{\sin }^{n+2}xdx=\frac{1}{n+1}{I}_{n+2}\end{array}$$从而有 $I_{n+2}=\frac{n+1}{n+2}{I}_n$, 故: $$\begin{array}{rl}{I}_{2p}& =\frac{(2p)!}{2^{2p}{(p!)}^2}\frac{\pi }{2}\\ I_{2p+1}& =\frac{2^{2p}{(p!)}^2}{(2p+1)!}\end{array}$$定义序列 $x_n=\sqrt{\frac{2n}{\pi }}{I}_n$, 则: $$\frac{x_{n+2}}{x_n}=\frac{n+1}{\sqrt{n(n+2)}}\ge 1,x_{2p}{x}_{2p+1}=\frac{2p}{2p+1}$$(10)故极限存在, 而 $x_n\le \sqrt{\frac{2n}{\pi }}{I}_{n-1}=x_{n-1}\sqrt{\frac{n}{n-1}}$, 保证奇偶极限相等.

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例 1.8 (Stirling 公式). 有: $$\underset{n\to +\infty }{\lim }\frac{e^{n}n!}{n^{n+\frac{1}{2}}}=\sqrt{2\pi }$$(11)

证明. 令 $a_m:=\frac{e^{n}n!}{n^{n+\frac{1}{2}}}$ 有: $$\ln \frac{a_n}{a_{n+1}}=\left(n+\frac{1}{2}\right)\ln \left(1+\frac{1}{n}\right)-1$$(12)在 $n\ge 3$ 时放缩, 有 $a_n\ge a_{n+1}$ 故极限存在.

利用 Wallis 积分已知: $$\frac{{(\frac{e^{n}p!}{p^{p+\frac{1}{2}}})}^2}{\frac{e^{2p}(2p)!}{(2p{)}^{2p+\frac{1}{2}}}\sqrt{2}}=\frac{{(p!)}^2{2}^{2p}}{(2p)!\sqrt{p}}\to \sqrt{\pi }$$(13)

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