2022-03-14 积分题

zedo2022/3/14大约 2 分钟约 724 字

题目

计算不定积分 $I = \int \frac{d x}{( x ^{2} + a ^{2} ) ^{2}}$ .

常规思路

直接凑微分得到

I = \int - \frac{1}{2 x} d (\frac{1}{x ^{2} + a ^{2}}) = - \frac{1}{2 x} \frac{1}{x ^{2} + a ^{2}} + \frac{1}{2} \int \frac{1}{x ^{2} + a ^{2}} d (\frac{1}{x})

其中

\int \frac{1}{x ^{2} + a ^{2}} d (\frac{1}{x}) = \frac{1}{a ^{3}} \int \frac{\frac{a ^{2}}{x ^{2}}}{1 + \frac{a ^{2}}{x ^{2}}} d (\frac{a}{x}) = \frac{1}{a ^{3}} \int 1 - \frac{1}{1 + \frac{a ^{2}}{x ^{2}}} d (\frac{a}{x}) = \frac{1}{a ^{2}} \frac{1}{x} - \frac{1}{a ^{3}} arctan \frac{a}{x} = \frac{1}{a ^{2}} \frac{1}{x} + \frac{1}{a ^{3}} arctan \frac{x}{a} - \frac{π}{2 a ^{3}}

因此

I = - \frac{1}{2 x ( x ^{2} + a ^{2} )} + \frac{1}{2 a ^{2} x} + \frac{1}{2 a ^{3}} arctan \frac{x}{a} = \frac{1}{2 a ^{2} x} (1 - \frac{a ^{2}}{x ^{2} + a ^{2}}) + \frac{1}{2 a ^{3}} arctan \frac{x}{a} = \frac{1}{2 a ^{2}} \frac{x}{x ^{2} + a ^{2}} + \frac{1}{2 a ^{3}} arctan \frac{x}{a} + C

看到 $x^{2} + a^{2}$ 很容易想到三角换元：

\int \frac{d x}{( x ^{2} + a ^{2} ) ^{2}} x = a t a n u \int \frac{a sec ^{2} u d u}{a ^{4} ( tan ^{2} u + 1 ) ^{2}} = \frac{1}{a ^{3}} \int \frac{d u}{sec ^{2} u} = \frac{1}{a ^{3}} \int cos^{2} u d u = \frac{1}{a ^{3}} \int \frac{1 + cos 2 u}{2} d u = \frac{u}{2 a ^{3}} + \frac{1}{4 a ^{3}} sin 2 u = \frac{1}{2 a ^{3}} arctan \frac{x}{a} + \frac{1}{4 a ^{3}} \frac{2 \frac{x}{a}}{1 + ( \frac{x}{a} ) ^{2}} = \frac{1}{2 a} arctan \frac{x}{a} + \frac{1}{2 a ^{2}} \frac{x}{a ^{2} + x ^{2}} + C

其中：

sin (2 arctan x) = 2 sin (arctan x) cos (arctan x) = 2 \cdot \frac{x}{1 + x ^{2}} \frac{1}{1 + x ^{2}} = \frac{2 x}{1 + x ^{2}}

注意到如下递推式：

I_{n + 1} = \frac{1}{2 a ^{2} n} [\frac{x}{( x ^{2} + a ^{2} ) ^{n}} + (2 n - 1) I_{n}]

其中 $I_{n} = \int \frac{d x}{( x ^{2} + a ^{2} ) ^{n}}$ ，且 $n ⩾ 1$ ，那么

I = I_{2} = \frac{1}{2 a ^{2}} [\frac{x}{x ^{2} + a ^{2}} + I_{1}] = \frac{1}{2 a ^{2}} [\frac{x}{x ^{2} + a ^{2}} + \int \frac{d x}{x ^{2} + a ^{2}}] = \frac{1}{2 a ^{2}} [\frac{x}{x ^{2} + a ^{2}} + \frac{1}{a} arctan \frac{x}{a}] + C

证明

直接分部积分

I_{n} = \int \frac{d x}{( x ^{2} + a ^{2} ) ^{n}} = \frac{x}{( x ^{2} + a ^{2} ) ^{n}} - \int x d [\frac{1}{( x ^{2} + a ^{2} ) ^{n}}] = \frac{x}{( x ^{2} + a ^{2} ) ^{n}} - \int x \frac{- n \cdot 2 x}{( x ^{2} + a ^{2} ) ^{n + 1}} d x = \frac{x}{( x ^{2} + a ^{2} ) ^{n}} + 2 n \int \frac{x ^{2} + a ^{2} - a ^{2}}{( x ^{2} + a ^{2} ) ^{n + 1}} d x = \frac{x}{( x ^{2} + a ^{2} ) ^{n}} + 2 n (I_{n} - a^{2} I_{n + 1})

化简即得：

I_{n + 1} = \frac{1}{2 a ^{2} n} [\frac{x}{( x ^{2} + a ^{2} ) ^{n}} + (2 n - 1) I_{n}]

实际上这也相当于三角函数 $n$ 次幂积分的递推式。