悠闲数学娱乐论坛(第2版)'s Archiver

青青子衿 发表于 2021-4-25 23:51

某高次有理分式的广义积分

[i=s] 本帖最后由 青青子衿 于 2021-5-23 08:24 编辑 [/i]

\begin{align*}
\int_{-\infty}^{+\infty}\left|\frac{Ax\mathrm{i}+B}{a(x\mathrm{i})^2+bx\mathrm{i}+c}\right|^2\mathrm{d}x=
\int_{-\infty}^{+\infty}\frac{A^{2}x^{2}+B^{2}}{b^{2}x^{2}+\left(c-ax^{2}\right)^{2}}\mathrm{d}x=\frac{aB^{2}+cA^{2}}{abc}\pi
\end{align*}

\begin{align*}
\color{black}{
\begin{split}   
\int_{-\infty}^{+\infty}\left|\frac{b_{0}+b_{1}x\mathrm{i}}{a_{0}+a_{1}x\mathrm{i}+a_{2}(x\mathrm{i})^2}\right|^2\mathrm{d}x&=   
\int_{-\infty}^{+\infty}\frac{{b_{0}\!}^{2}+{b_{1}}\!^{2}x^{2}}{{a_1}\!^{2}x^{2}+\left(a_0-a_2x^{2}\right)^{2}}\mathrm{d}x\\
\\
&=\frac{a_0{b_1}\!^{2}+a_{2}{b_0}\!^{2}}{a_{0}a_{1}a_{2}}\pi\\
\\
\int_{-\infty}^{+\infty}\left|\frac{b_{0}+b_{1}x\mathrm{i}+b_{2}(x\mathrm{i})^2}{a_{0}+a_{1}x\mathrm{i}+a_{2}(x\mathrm{i})^2+a_{3}(x\mathrm{i})^3}\right|^2\mathrm{d}x&=   
\int_{-\infty}^{+\infty}\frac{{b_1}\!^{2}x^{2}+\left(b_0-b_2x^{2}\right)^{2}}{(a_{0} - a_{2}x^2)^2 + x^2 (a_{1}- a_{3} x^2)^2}\mathrm{d}x\\
\\
&=\frac{a_{2}a_{3}b_{0}^{2}+a_{0}a_{3}b_{1}^{2}+a_{0}a_{1}b_{2}^{2}-2a_{0}a_{3}b_{0}b_{2}}{a_{0}a_{3}\left(a_{1}a_{2}-a_{0}a_{3}\right)}\pi\\
\end{split}}
\end{align*}

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