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[分享] 一道二重瑕积分

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发表于 2019-1-6 16:47:39 | 显示全部楼层 |阅读模式

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本帖最后由 葡萄糖 于 2019-1-6 17:58 编辑

\[ \color{black}{\int_0^1\int_0^1\frac{\sqrt{x+y\,}}{x^2+y^2}{\rm\,d}x{\rm\,d}y} \]
kastin 发表于 2014-7-26 15:25
这种类型的不定积分,大概在2年前我就很熟悉了,下面我将展示求这类有理分式不定积分过程中"拆"和“凑”的 ...


\begin{align*}  
&&\int_0^1\frac{\sqrt{x+y\,}}{x^2+y^2}{\rm\,d}x&
=\int_0^1\frac{u}{\left(u^2-y\right)^2+y^2}\left(2u\right){\rm\,d}u\\
&&&=2\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\frac{u^2}{\left(u^2-y\right)^2+y^2}{\rm\,d}u\\
&&&=\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\frac{\left(u^2+\sqrt{2}y\,\right)+\left(u^2-\sqrt{2}y\,\right)}{u^4+2y^2-2u^2y}{\rm\,d}u\\
&&&=\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\left(\frac{1+\frac{\sqrt{2}\,y}{u^2}}{u^2+\frac{2\,y^2}{u^2}-2y}\right)\!{\rm\,d}u
+\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\left(\frac{1-\frac{\sqrt{2}\,y}{u^2}}{u^2+\frac{2\,y^2}{u^2}-2y}\right)\!{\rm\,d}u\\
&&&=\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\frac{{\rm\,d}\left(u-\frac{\sqrt{2}\,y}{u}\right)}{\left(u-\frac{\sqrt{2}\,y}{u}\right)^2+2\left(\sqrt{2\,}-1\right)y}  
+\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\frac{{\rm\,d}\left(u+\frac{\sqrt{2}\,y}{u}\right)}{\left(u+\frac{\sqrt{2}\,y}{u}\right)^2-2\left(\sqrt{2}+1\right)y}\\
&&&=+\end{align*}
\begin{align*}
\int \frac{{\rm\,d}z}{z^2+{c_1}^2}&=\frac{1}{c_1}\arctan\frac{z}{c_1}+C\\  
\int \frac{{\rm\,d}z}{z^2-{c_2}^2}&=\frac{1}{2c_2}\ln\left|\frac{z-c_2}{z+c_2}\right|+C
\end{align*}

  1. [quote][size=2][url=forum.php?mod=redirect&goto=findpost&pid=54898&ptid=5715][color=#999999]kastin 发表于 2014-7-26 15:25[/color][/url][/size]
  2. 这种类型的不定积分,大概在2年前我就很熟悉了,下面我将展示求这类有理分式不定积分过程中"拆"和“凑”的 ...[/quote]
复制代码

葡萄糖 发表于 2019-1-6 16:46
\[2\int_0^1\frac{u^2}{\left(u^2-y\right)^2+y^2}{\rm\,d}u\] ...


\begin{align*}  
&&\int_0^1\frac{\sqrt{x\,}}{\left(x-y\right)^2+y^2}{\rm\,d}x&
=\int_0^1\frac{u}{\left(u^2-y\right)^2+y^2}\left(2u\right){\rm\,d}u\\
&&&=2\int_0^1\frac{u^2}{\left(u^2-y\right)^2+y^2}{\rm\,d}u\\
&&&=\int_0^1\frac{\left(u^2+\sqrt{2}y\,\right)+\left(u^2-\sqrt{2}y\,\right)}{u^4+2y^2-2u^2y}{\rm\,d}u\\
&&&=\int_0^1\left(\frac{1+\frac{\sqrt{2}\,y}{u^2}}{u^2+\frac{2\,y^2}{u^2}-2y}\right)\!{\rm\,d}u
+\int_0^1\left(\frac{1-\frac{\sqrt{2}\,y}{u^2}}{u^2+\frac{2\,y^2}{u^2}-2y}\right)\!{\rm\,d}u\\
&&&=\int_0^1\frac{{\rm\,d}\left(u-\frac{\sqrt{2}\,y}{u}\right)}{\left(u-\frac{\sqrt{2}\,y}{u}\right)^2+2\left(\sqrt{2\,}-1\right)y}   
+\int_0^1\frac{{\rm\,d}\left(u+\frac{\sqrt{2}\,y}{u}\right)}{\left(u+\frac{\sqrt{2}\,y}{u}\right)^2-2\left(\sqrt{2}+1\right)y}\\
&&&=\frac{1}{\sqrt{2\left(\sqrt{2}-1\right)y}}\arctan\left(\frac{\sqrt{2\left(\sqrt{2}-1\right)y}}{\sqrt{2}y-1}\right)
+\frac{1}{2\sqrt{2\left(\sqrt{2}+1\right)y}}\ln\left|\frac{1+\sqrt{2}y-\sqrt{2\left(\sqrt{2}+1\right)y}}{1+\sqrt{2}y+\sqrt{2\left(\sqrt{2}+1\right)y}}\right|
\end{align*}
\begin{align*}
\int \frac{{\rm\,d}z}{z^2+{c_1}^2}&=\frac{1}{c_1}\arctan\frac{z}{c_1}+C\\  
\int \frac{{\rm\,d}z}{z^2-{c_2}^2}&=\frac{1}{2c_2}\ln\left|\frac{z-c_2}{z+c_2}\right|+C
\end{align*}
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-1-9 10:06:40 | 显示全部楼层
  1. NIntegrate[Sqrt[x + y]/(x^2 + y^2), {x, 0, 1}, {y, 0, 1},
  2. AccuracyGoal -> 10, WorkingPrecision -> 20]
复制代码


3.7537956645732992068
数值解万岁!
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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