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自然对数函数

— The Natural Logarithm —

Haifeng Xu

(hfxu@yzu.edu.cn)

This slide is based on Selwyn Hollis's work, which can be download from his website.

目录

自然对数函数的定义

$\ln x$ 函数定义为

\[\ln x=\int_1^x\frac{1}{t}dt,\quad\forall\ x>0.\]

这等价于

\[ \left\{ \begin{array}{rcl} \frac{d}{dx}\ln x&=&\frac{1}{x},\quad\forall\ x>0,\\ \\ \ln 1 &=& 0. \end{array} \right. \]

代数性质

极限

$\lim\limits_{x\rightarrow+\infty}\ln x=+\infty,\quad$ 这是因为 $\ \ln(x^p)=p\ln x,\quad$ 从而 $\ \ln(2^x)=x\ln 2$.

$\lim\limits_{x\rightarrow 0^+}\ln x$$\ =$$\lim\limits_{x\rightarrow +\infty}\ln(1/x)$$\ =$$\lim\limits_{x\rightarrow +\infty}(-\ln x)$$\ =-\infty$.

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定理: \[ \lim_{x\rightarrow +\infty}\frac{\ln x}{x^r}=0,\quad\forall\ r>0. \]

证明: 选取 $p\in(0,1),\ $ 且 $\ p > 1-r.\ $ 则对于 $\ x > 1$,

$\displaystyle\ln x=\int_1^x\frac{1}{t}dt$ $\displaystyle\ < \int_1^x\frac{1}{t^p}dt $ $\displaystyle\ =\frac{1}{1-p}t^{1-p}\bigg|_1^x $ $\displaystyle\ =\frac{x^{1-p}-1}{1-p},$

故

$\displaystyle 0 < \frac{\ln x}{x^r}$ $\displaystyle < \frac{x^{1-p-r}-x^{-r}}{1-p} $ $\displaystyle\rightarrow 0, \quad (\mbox{当} x\rightarrow +\infty)$

由夹逼原理知定理成立.

$x^r\ln x$ 的性质

定理: \[ \lim_{x\rightarrow +\infty}\frac{\ln x}{x^r}=0,\quad\forall\ r>0. \]

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推论: \[ \lim_{x\rightarrow 0^{+}}x^r\ln x=0,\quad\forall\ r>0. \]

证明:

$\displaystyle\lim_{x\rightarrow 0^+}x^r\ln x$ $\displaystyle\ =\,\lim_{t\rightarrow +\infty}(\frac{1}{t})^r\ln\frac{1}{t}$$\displaystyle\ =\,-\lim_{t\rightarrow +\infty}\frac{\ln t}{t^r}$$\displaystyle\ =\,0.$

$x^r\ln x$ 的图像

$\displaystyle y=\frac{\ln x}{x^2}$,   $\displaystyle y=\frac{\ln x}{\sqrt{x}}$,   $\displaystyle y=\sqrt{x}\ln x$,   $\displaystyle y=x^2\ln x$.
图片

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函数 $\ln |x|$

图像

性质

\[ \frac{d}{dx}\ln |x|=\frac{1}{x},\quad\forall\ x\neq 0. \]

\[ \int\frac{1}{x}dx=\ln |x|+C. \]

$x^n$ 的不定积分 \[ \int x^n dx= \begin{cases} \frac{1}{n+1}x^{n+1}+C,& 若 n\neq -1,\\ \ln |x|+C,& 若 n=-1. \end{cases} \]

与 $\ln x$ 有关的不定积分（例子）

例 1. $\displaystyle\int\frac{x+1}{x^2}dx$$\displaystyle\ =\int(\frac{1}{x}+\frac{1}{x^2})dx$ $\displaystyle\ =\ln |x|-\frac{1}{x}+C.$

例 2. $\displaystyle\int\frac{x}{x^2+1}dx$

令 $u=x^2+1,\ $ 则 $du=2xdx,\ $ 于是

$\displaystyle\int\frac{x}{x^2+1}dx$ $\displaystyle\ =\,\frac{1}{2}\int\frac{1}{u}du$ $\displaystyle\ =\,\frac{1}{2}\ln|u|+C$ $\displaystyle\ =\,\frac{1}{2}\ln|x^2+1|+C.$

例 3.

$\displaystyle\int\tan xdx$ $\displaystyle\ =\,\int\frac{\sin x}{\cos x}dx$ $\displaystyle\ =\,\int\frac{d(-\cos x)}{\cos x}$ $\displaystyle\ =\,-\ln |\cos x|+C$ $\displaystyle\ =\,\ln |\sec x|+C.$

例 4.

$\displaystyle\int\frac{1}{\sqrt{x}(1+\sqrt{x})}dx$ $\displaystyle\ =\,\int\frac{2d\sqrt{x}}{1+\sqrt{x}}$ $\displaystyle\ =\,2\int\frac{d(1+\sqrt{x})}{1+\sqrt{x}}$ $\displaystyle\ =\,2\ln(1+\sqrt{x})+C.$

形如 $\frac{u'}{u}$ 函数的不定积分

\[ \frac{d}{dx}\ln |u(x)|=\frac{u'(x)}{u(x)}, \]

\[ \int\frac{u'(x)}{u(x)}dx=\ln |u(x)|+C. \]

例 1.

$\displaystyle\int\frac{\cos x}{\sin x}dx$ $\displaystyle\ =\,\int\frac{d\sin x}{\sin x}$ $\displaystyle\ =\,\ln |\sin x|+C.$

例 2.

$\displaystyle\int\frac{\cos x}{\sin^2 x}dx$ $\displaystyle\ =\,\int\frac{d\sin x}{\sin^2 x}$ $\displaystyle\ =\,\int\frac{du}{u^2}$ $\displaystyle\ =\,-\frac{1}{u}+C$ $\displaystyle\ =\,-\frac{1}{\sin x}+C.$

幂指函数的导数/微分

例. 求 $y=x^{-x}$ 的导数

解. $\displaystyle y = x^{-x},$ 两边取自然对数, $\displaystyle\ln y=\ln x^{-x}=-x\ln x$. 两边对 $x$ 求导,

$\displaystyle\frac{1}{y}\frac{dy}{dx}$ $\displaystyle\ =\,-(1\cdot\ln x+x\cdot\frac{1}{x})$

从而

$\displaystyle\frac{dy}{dx}$ $\displaystyle\ =\,-(\ln x+1)y$ $\displaystyle\ =\,-(\ln x+1)x^{-x}.$

法二

$\displaystyle (x^{-x})'$ $\displaystyle\ =\,(e^{-x\ln x})'$ $\displaystyle\ =\,e^{-x\ln x}\cdot(-x\ln x)'$ $\displaystyle\ =\,x^{-x}\cdot(-1)(1\cdot\ln x+x\cdot\frac{1}{x})$ $\displaystyle\ =\,-x^{-x}(\ln x+1).$

End

Thanks very much!