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Lecture07.pdf第3页
$P_1,P_2,P_3$是$x$的多项式,$P_1,P_2$互素,则可以求出两个多项式$A_1,A_2$使得$A_{1} P_{1}+A_{2} P_{2}=P_{3}$
为了求$$\int \frac{P(x)}{Q(x)} d x$$
设$Q(x)=Q_{1}(x) Q_{2}(x)^{2} Q_{3}(x)^{3} \ldots Q_{n}(x)^{n}$,其中$Q_1,Q_2,⋯,Q_n$是无重根(squarefree)的互素的多项式.
可以求出$B,A_1$使$$B Q_{1}+A_{1} Q_{2}^{2} Q_{3}^{3} \ldots Q_{n}^{n}=P$$
则$$R(x)=\frac{P}{Q}=\frac{A_{1}}{Q_{1}}+\frac{B}{Q_{2}^{2} Q_{3}^{3} \ldots Q_{n}^{n}}$$
重复这个过程得到$$R(x)=\frac{A_{1}}{Q_{1}}+\frac{A_{2}}{Q_{2}^{2}}+\frac{A_{3}}{Q_{3}^{3}}+\cdots+\frac{A_{n}}{Q_{n}^{n}}$$
所以有理函数可以拆成分母无重根的有理函数之和.
因为$Q$无重根,$Q,Q'$互素,所以存在$C,D$使得$$C Q+D Q^{\prime}=A$$
所以\begin{aligned} \int \frac A{Q^n} d x &=\int \frac{C Q+D Q^{\prime}}{Q^{n}} d x \\ &=\int \frac{C}{Q^{n-1}} d x-\frac{1}{n-1} \int D \frac{d}{d x}\left(\frac{1}{Q^{n-1}}\right) d x \\ &=-\frac{D}{(n-1) Q^{n-1}}+\frac{1}{n-1} \int \frac{D^{\prime}}{Q^{n-1}} d x+\int \frac{C}{Q^{n-1}} d x \\ &=-\frac{D}{(n-1) Q^{n-1}}+\int \frac{E}{Q^{n-1}} d x \end{aligned}
其中$$E=C+\frac{D^{\prime}}{n-1}$$所以分母中$x$的次数下降了$\deg Q$,我们得到了一个类似的积分,我们可以继续进行下去,最终得到
$$\int \frac{A}{Q^{n}} d x=R_{n}(x)+\int \frac{S}{Q} d x$$
其中$R_n$是一个有理函数,$S$是一个多项式,而$\int \frac{E}{Q^{n-1}} d x$中没有有理函数部分,因为$Q$是无重根的.因此,$\int R(x)dx$的有理函数部分为$$R_{2}(x)+R_{3}(x)+\cdots+R_{t}(x)$$
Example 9.2.1 为了积分$$\int \frac{4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}-3}{\left(x^{7}-x+1\right)^{2}} d x$$
取$P_{1}=x^{7}-x+1, P_{2}=P_{1}^{\prime}=7 x^{6}-1, P_{3}=4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}$
设$C\left(x^{7}-x+1\right)+D\left(7 x^{6}-1\right)=4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}-3$
所以我们可以取$C$的次数不超过 5，$D$的次数不超过 6。然后我们得到一个包含 13 个未知数的 13 个方程的组，我们可以求解。在这种情况下，我们发现：
\begin{aligned}
&C=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \\
&D=b_{6} x^{6}+b_{5} x^{5}+b_{4} x^{4}+b_{3} x^{3}+b_{2} x^{2}+b_{1} x+b_{0} \\
&C \times\left(x^{7}-x+1\right)+D \times\left(7 x^{6}-1\right)=4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}-3 \\
&\left(a_{5}+7 b_{6}\right) x^{12}+\left(a_{4}+7 b_{5}\right) x^{11}+\left(a_{3}+7 b_{4}\right) x^{10}+\left(a_{2}+7 b_{3}\right) x^{9}+\left(a_{1}+7 b_{2}\right) x^{8}+\left(a_{0}+7 b_{1}\right) x^{7}+ \\
&\quad\left(7 b_{0}-b_{6}-a_{5}\right) x^{6}+\left(a_{5}-b_{5}-a_{4}\right) x^{5}+\left(a_{4}-b_{4}-a_{3}\right) x^{4}+\left(a_{3}-b_{3}-a_{2}\right) x^{3}+\left(a_{2}-b_{2}-a_{1}\right) x^{2}+ \\
&\quad\left(a_{1}-b_{1}-a_{0}\right) x+\left(a_{0}-b_{0}\right)=4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}-3
\end{aligned}
得出方程组
\begin{array}{c}a_{5}+7 b_{6}=0 \\ a_{4}+7 b_{5}=0 \\ a_{3}+7 b_{4}=0 \\ a_{2}+7 b_{3}=4 \\ a_{1}+7 b_{2}=0 \\ a_{0}+7 b_{1}=0 \\ 7 b_{0}-b_{6}-a_{5}=21 \\ a_{5}-b_{5}-a_{4}=0 \\ a_{4}-b_{4}-a_{3}=0 \\ a_{3}-b_{3}-a_{2}=2 \\ a_{2}-b_{2}-a_{1}=-3 \\ a_{1}-b_{1}-a_{0}=0 \\ a_{0}-b_{0}=-3\end{array}
解得
\begin{array}{l}a_{5}=a_{4}=a_{2}=a_{1}=a_{0}=0 \\ a_{3}=-3 \\ b_{6}=b_{5}=b_{4}=b_{2}=b_{1}=0 \\ b_{3}=1, \quad b_{0}=3\end{array}
于是
$$C=-3 x^{2}, \quad D=x^{3}+3$$
根据Hermite算法有
$$\int \frac{4 x^{9}+21 x^{6}+2 x^{3}-3 x^{2}-3}{\left(x^{7}-x+1\right)^{2}} d x=-\frac{x^{3}+3}{x^{7}-x+1}$$$$S=-3 x^{2}+\frac{d}{d x}\left(x^{3}+3\right)=0$$ |
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