Section02_基本理论

积分基本定理

不定积分与定积分的不同

1. $\displaystyle \int f(x)\cdot dx \neq \int f(t) \cdot dt$
2. $\displaystyle \int_{a}^{b} f(x) \cdot dx = \int_{a}^{b} f(t)\cdot dt = \int_{a}^{b} f(u)\cdot du$

积分上限函数

• 对于曲线$\text{L: }y=f(x)\quad x\in[a,b]$，$\forall\ x\in[a,b]$ $$\Phi(x) \triangleq\int_{a}^{x}f(t)\cdot dt$$

Notes

1. $\displaystyle \int_{a}^{x} f(x)\cdot dx$ 表达式中的 $x$ 和上限的 $x$ 不同
2. $\displaystyle \int_{a}^{x}f(x,t)\cdot dt$ 表达式中的 $x$ 和上限的 $x$ 相同
3. 若 $f(x)\in C[a,b]$，则 $\displaystyle \int_{a}^{x}f(t)\cdot dt$ 可导
4. 若 $f(x)\in C[a,b]$，则 $f(x)$ 存在原函数
5. $\displaystyle\frac{d}{dx}\int_{a}^{\phi(x)}f(x)\cdot dx = f[\phi(x)]\cdot \phi'(x)$
6. $\displaystyle \frac{d}{dx}\int_{\phi_{1}(x)}^{\phi_{2}(x)}f(x)\cdot dx = f[\phi_{2}(x)]\cdot \phi_{2}'(x) - f[\phi_{1}(x)]\cdot \phi_{1}'(x)$

定理1

• 设 $f(x)\in C[a,b]$ 或除去有限个第一类间断点外处处连续，$\displaystyle \Phi(x) = \int_{a}^{x}f(t)\cdot dt$，则 $\Phi'(x) = f(x)$

Proof $$\begin{array}{ll} \because & \displaystyle \Phi(x) = \int_{a}^{x}f(t)\cdot dt \\ \therefore &\displaystyle \Delta \Phi = \Phi(x+\Delta x) - \Phi(x) = \int_{a}^{x+\Delta x}f(t)\cdot dt - \int_{a}^{x}f(t)\cdot dt\\ & = \displaystyle \int_{x}^{x+\Delta x} f(t)\cdot dt \\ \because & f(x)\in C[a,b] \text{根据定积分中值定理} \\ \therefore & \exists\ \zeta\in(x,x+\Delta x), f(\zeta)\Delta x = \Delta\Phi \\ \therefore & \displaystyle \lim_{\Delta x\to 0} f(\zeta) = f(x) = \lim_{\Delta x\to 0}\frac{\Delta \Phi}{\Delta x} = \Phi'(x) \\ & \Phi'(x) = f(x)\end{array}$$

定理二 牛顿-莱布尼茨公式

• 设 $f(x)\in C[a,b]$，$F(x)$ 为 $f(x)$ 的一个原函数，则：$\displaystyle \int_{a}^{b}f(x)\cdot dx = F(b)-F(a)$

Proof $$\begin{array}{ll} & \displaystyle\text{设}\Phi(x) = \int_{a}^{x}f(x)\cdot dx \\ \because & \Phi'(x) = f(x) \\ \therefore & \Phi(x) \text{为}f(x)\text{的原函数} \\ \because & F(x)\text{也为}f(x)\text{的一个原函数} \\ \therefore & F(b)-\Phi(b) = C_{0}, F(a)-\Phi(a) = C_{0} \\ & \displaystyle\text{而}\Phi(a) = \int_{a}^{a}f(x)\cdot dx = 0 \\ \therefore & F(b) - F(a) = \Phi(b) \\ & \displaystyle\text{即} \int_{a}^{b}f(x)\cdot dx = F(b) - F(a)\end{array}$$

例题

• 例1 $\displaystyle \lim_{x\to 0}\frac{\int_{0}^{x}e^{-t^{2}}\cos t\cdot dt - x}{x^{3}}$ $$\begin{split} & \lim_{x\to 0}\frac{\int_{0}^{x}e^{-t^{2}}\cos t\cdot dt - x}{x^{3}} = \lim_{x\to 0} \frac{e^{-x^{2}}\cos x - 1}{3x^{2}} \\ = & \lim_{x\to 0}\frac{e^{-x^{2}}(\cos x - 1)+(e^{-x^{2}}-1)}{3x^{2}} \\ = & \lim_{x\to 0}e^{-x^{2}}\frac{\cos x -1}{3x^{2}} +\lim_{x\to 0}\frac{e^{-x^{2}}-1}{3x^{2}} \\ = & \lim_{x\to 0}\frac{-\frac{1}{2}x^{2}}{3x^{2}} + \lim_{x\to 0}\frac{-x^{2}}{3x^{2}} \\ =& -\frac{1}{2} \end{split}$$
• 例2 $f(x)$连续，$f(0)=0,f'(0)=6$，求$\displaystyle \lim_{x\to 0}\frac{\int_{0}^{x}(x-t)f(t)\cdot dt}{x^3}$ $$\begin{split} & \lim_{x\to 0}\frac{\int_{0}^{x}(x-t)f(t)\cdot dt}{x^{3}}\\ = & \lim_{x\to 0} \frac{x \int_{0}^{x}f(t)\cdot dt-\int_{0}^{x}tf(t)\cdot dt}{x^{3}} \\ = & \lim_{x\to 0} \frac{\int_{0}^{x}f(t)\cdot dt + xf(x)-xf(x)}{3x^{2}} \\ = & \lim_{x\to 0} \frac{f(x)}{6x} = \frac{1}{6}\lim_{x\to 0}\frac{f(x)- f(0)}{x-0} \\ = & \frac{1}{6}f'(0) = 1 \end{split}$$

定积分的特殊性质

对称性质

• $f(x)\in C[-a,a]$，则 $$\int_{-a}^{a}f(x)\cdot dx = \int_{0}^{a}\bigg[f(x)+f(-x)\bigg]\cdot dx$$

  Proof $$\begin{array}{ll}& \displaystyle \text{左式} = \int_{-a}^{0}f(x)\cdot dx + \int_{0}^{a}f(x)\cdot dx\\ \because & \displaystyle \int_{-a}^{0} f(x)\cdot dx \xlongequal{x=-t} \int_{a}^{0}f(-t)\cdot d(-t) \\ &\displaystyle = -\int_{a}^{0}f(-t)\cdot dt = \int_{0}^{a}f(-x)\cdot dx\\ \therefore &\displaystyle \text{左式} = \int_{0}^{a}f(-x)\cdot dx + \int_{0}^{a}f(x)\cdot dx \\&\displaystyle = \int_{0}^{a}\big[f(x)+f(-x)\big]\cdot dx \\\end{array}$$

例题

• 例1 $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{dx}{1+\sin x}$ $$\begin{split} &\displaystyle \text{原式} = \int_{0}^{\frac{\pi}{4}}\bigg[\frac{1}{1+\sin x} + \frac{1}{1-\sin x}\bigg]\cdot dx \\ = & \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{2}{1-\sin^{2}x} \cdot dx = 2\int_{0}^{\frac{\pi}{4}}\sec^{2}x\cdot dx \\ = &\displaystyle 2\tan x\vert_{0}^{\frac{\pi}{4}} = 2 \end{split}$$

三角函数

性质一 $\displaystyle \int_{0}^{\frac{\pi}{2}} f(\sin x)\cdot dx = \int_{0}^{\frac{\pi}{2}} f(\cos x)\cdot dx$

• 若 $f(x)\in C[0,1]$，则 $$\int_{0}^{\frac{\pi}{2}} f(\sin x)\cdot dx = \int_{0}^{\frac{\pi}{2}} f(\cos x)\cdot dx$$

  Proof $$\begin{array}{ll} &\displaystyle \int_{0}^{\frac{\pi}{2}}f(\sin x)\cdot dx \xlongequal{t = \frac{\pi}{2}- x} \int_{\frac{\pi}{2}}^{0}f(\cos t)\cdot d\bigg(\frac{\pi}{2}-t\bigg)\\ = & \displaystyle -\int_{\frac{\pi}{2}}^{0}f(\cos t)\cdot dt = \int_{0}^{\frac{\pi}{2}}f(\cos x)\cdot dx \\\end{array}$$

• 特别地 $\displaystyle I_{n}\triangleq \int_{0}^{\frac{\pi}{2}}\sin^{n} x\cdot dx = \int_{0}^{\frac{\pi}{2}}\cos^{n} x\cdot dx$ $$\begin{cases} I_{n} = \frac{n-1}{n}I_{n-2} \\ I_{1} = 1 \\ I_{0} = \frac{\pi}{2} \end{cases}$$

  Proof $I_{n}= \frac{n-1}{n}I_{n-2}$ $$\begin{array}{ll} & \displaystyle 令\mathcal{I}_{n} = \int_{}^{}\sin^{n}x\cdot dx = \int_{}^{}\cos^{n}x\cdot dx \\ & \displaystyle \mathcal{I}_{n} = \int_{}^{} \sin^{n}\cdot dx = -\int_{}^{} \sin^{n-1}\cdot d(\cos x) \\ &= \displaystyle- \cos x\sin ^{n-1} x + (n-1)\int_{}^{}\cos x \cdot \sin^{n-2}x \cos x\cdot dx \\ &= \displaystyle - \cos x\sin^{n-1} x + (n-1)\int_{}^{}(1-\sin^{2}x)\sin^{n-2}x\cdot dx \\ &= -\cos x\sin ^{n-1}x + (n-1)\mathcal{I}_{n-2} - (n-1)\mathcal{I}_{n} \\ \therefore & n\mathcal{I}_{n} = -\cos x\sin ^{n-1}x +(n-1)\mathcal{I}_{n-2}\\ & \mathcal{I}_{n} = -\frac{1}{n}\cos x\sin^{n-1}x+\frac{n-1}{n}\mathcal{I}_{n-2}\\ & \displaystyle\text{令}I_{n} = \int_{0}^{\frac{\pi}{2}}\sin ^{n}x\cdot dx = \int_{0}^{\frac{\pi}{2}}\cos^{n}x\cdot dx\\ \therefore & \displaystyle I_{n} = -\frac{1}{n}\cos x\sin^{n-1}x\vert_{0}^{\frac{\pi}{2}} + \frac{n-1}{n}I_{n-2}\\ \because & \cos x\sin^{n-1}x\vert_{0}^{\frac{\pi}{2}} = 0 \\ \therefore & I_{n} = \frac{n-1}{n}I_{n-2} \\\end{array}$$

• 例1 $\displaystyle\int_{0}^{1}\frac{dx}{x+\sqrt{1-x^{2}}}$ $$\begin{array}{ll} \because&\displaystyle \int_{0}^{1}\frac{dx}{x+\sqrt{1-x^{2}}} \xlongequal{x=\sin t} \int_{0}^{\frac{\pi}{2}} \frac{\cos t}{\sin t+\cos t}\cdot dt = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{\cos t +\sin t}\cdot dt \\ \therefore & \displaystyle 2 \int_{0}^{\frac{\pi}{2}}\frac{\cos t}{\sin t + \cos t}\cdot dt = \int_{0}^{\frac{\pi}{2}}\frac{\sin t + \cos t}{\sin t + \cos t}\cdot dt = \int_{0}^{\frac{\pi}{2}}1\cdot dx = \frac{\pi}{2} \\ \therefore & \text{原式} = \frac{\pi}{4} \end{array}$$

• 例2 $\displaystyle I = \int_{0}^{2}x^{2}\sqrt{2x-x^{2}}\cdot dx$ $$\begin{split} & \displaystyle \int_{0}^{2}x^{2} \sqrt{2x-x^{2}}\cdot dx = \int_{0}^{2} [1+(x-1)]^{2}\sqrt{1-(x-1)^{2}}\cdot d(x-1) \\ = & \displaystyle \int_{-1}^{1}(1+x)^{2}\sqrt{1-x^{2}}\cdot dx \xlongequal{x= \sin t} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin t)^{2}\cos t \cdot d(\sin t) \\ = & \displaystyle \int_{0}^{\frac{\pi}{2}}\big[\cos^{2} t(2 + 2\sin^{2}t)\big]\cdot dt = 2 \int_{0}^{\frac{\pi}{2}}\cos^{2}t(2-\cos^{2}t) \\ = & 2(2I_{2}-I_{4}) = 2(2\cdot \frac{1}{2}\cdot\frac{\pi}{2}-\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi}{2}) = \pi - \frac{3}{8}\pi\\ = & \frac{5}{8}\pi \end{split}$$
• 例3
  1. $f,g\in C[-a,a], f(-x)+f(x) = A,g(-x)=g(x)$，证明 $\displaystyle \int_{-a}^{a}f(x)g(x)\cdot dx = A\int_{0}^{a}g(x)\cdot dx$ $$\begin{array}{ll} \because & \int_{-a}^{a}f(x)g(x)\cdot dx = \int_{0}^{a}[f(x)g(x)+f(-x)g(-x)]\cdot dx \\ & \text{且}f(x)+f(-x)=A, g(x) = g(-x) \\ \therefore & \text{原式} = \int_{0}^{a}g(x)[f(x)+f(-x)]\cdot dx = \int_{0}^{a}Ag(x)\cdot dx \\ &= A \int_{0}^{a}g(x)\cdot dx \end{array}$$
  2. 求$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan e^{x}\sin^{2}x\cdot dx$ $$\begin{array}{ll} & \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \int_{0}^{\frac{\pi}{2}}\sin^{2}x(\arctan e^{x} + \arctan e^{-x})\cdot dx \\ & \text{令} f(x) = \arctan e^{x} + \arctan e^{-x} \\ \therefore & f'(x) = \frac{e^{x}}{1+e^{2x}} + \frac{-e^{-x}}{1+e^{-2x}} = \frac{e^{x}}{1+e^{2x}} - \frac{e^{x}}{e^{2x}+1} = 0\\ \therefore & \arctan e^{x} + \arctan e^{-x}\text{为常数}\frac{\pi}{2} \\ \therefore & \text{原式} = \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\sin^{2}x\cdot dx = \frac{\pi}{2}I_{2} = \frac{\pi}{2}\cdot \frac{1}{2}\cdot \frac{\pi}{2} = \frac{\pi^{2}}{8} \end{array}$$

性质二 $\displaystyle \int_{0}^{\pi}f(\sin x)\cdot dx = 2\int_{0}^{\frac{\pi}{2}}f(\sin x)\cdot dx$

Proof $$\begin{array}{ll}\because & \int_{0}^{\pi}f(\sin x)\cdot dx = \int_{0}^{\frac{\pi}{2}} f(\sin x)\cdot dx+ \int_{\frac{\pi}{2}}^{\pi}f(\sin x)\cdot dx \\& \int_{\frac{\pi}{2}}^{\pi} f(\sin x)\cdot dx \xlongequal{x=t+\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}}f(\sin (t+\frac{\pi}{2})) \cdot dt = \int_{0}^{\frac{\pi}{2}}f(\cos t)\cdot dt\\\therefore & \text{原式} = \int_{0}^{\frac{\pi}{2}}f(\sin x)\cdot dx + \int_{0}^{\frac{\pi}{2}}f(\cos x)\cdot dx \\& = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x)\cdot dx\end{array}$$

• 例1 $\displaystyle I =\int_{-\pi}^{\pi}\frac{\sin^{2}x}{1+e^{x}}\cdot dx$ $$\begin{split} & I = \int_{0}^{\pi}\sin^{2}{x}(\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}})\cdot dx \\ = & \int_{0}^{\pi}\sin^{2}x(\frac{1}{1+e^{x}}+ \frac{e^{x}}{e^{x}+1})\cdot dx \\ = & \int_{0}^{\pi}\sin^{2}x\cdot dx \\ = & 2 \int_{0}^{\frac{\pi}{2}}\sin^{2} x\cdot dx = 2I_{2} \\ = & 2 \cdot \frac{1}{2}\cdot \frac{\pi}{2} = \frac{\pi}{2} \end{split}$$ Notes
• $\cos x$无此性质，而 $\vert \cos x\vert, \cos^{2n} x$ 则有此性质

性质三 $\displaystyle \int_{0}^{\pi}xf(\sin x)\cdot dx= \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)\cdot dx$

Proof $$\begin{array}{ll}\because & I = \int_{0}^{\pi}xf(\sin x)\cdot dx \xlongequal{x+t = \pi} \int_{\pi}^{0}(\pi -t)f[\sin(\pi-t)] \cdot d(\pi - t) \\& = \int_{0}^{\pi}(\pi-x)f(\sin x)\cdot dx = \pi \int_{0}^{\pi}f(\sin x)\cdot dx - I \\\therefore & 2I = \pi \int_{0}^{\pi}f(\sin x)\cdot dx \\& I = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)\cdot dx\end{array}$$

• 例1 $\displaystyle I = \int_{0}^{\pi}x\sin^{2} x\cdot dx$ $$\begin{split} &I = \frac{\pi}{2}\int_{0}^{\pi} \sin ^{2} x\cdot dx\\ = & \pi \int_{0}^{\frac{\pi}{2}}\sin^{2} x \cdot dx = \pi I_{2} = \pi\times \frac{1}{2}\times \frac{\pi}{2}\\ = &\frac{\pi^{2}}{4} \end{split}$$

周期函数

• $f(x)$连续且以 $T$ 为周期
  1. $\displaystyle \int_{a}^{a+T}f(x)\cdot dx = \int_{0}^{T}f(x)\cdot dx$
  2. $\displaystyle \int_{0}^{nT}f(x)\cdot dx = n\int_{0}^{T}f(x)\cdot dx$