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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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\[ \color{black}{\int_0^1\int_0^1\frac{\sqrt{x+y\,}}{x^2+y^2}{\rm\,d}x{\rm\,d}y} \]
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}\\  
&&&=
\color{red}{\frac{1}{\sqrt{2\left(\sqrt{2}-1\right)y}}\arctan\left(\frac{\sqrt{\frac{1+y}{y}}-\sqrt{\frac{2y}{1+y}}}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)+\frac{1}{\sqrt{2\left(\sqrt{2}-1\right)y}}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\right)}\\
&&&\mathrel{\phantom{=}}
+
\color{red}{\frac{1}{2\sqrt{2\left(\sqrt{2}+1\right)y}}\ln\left|\frac{1+\left(\sqrt{2}+1\right)y-\sqrt{2\left(\sqrt{2}+1\right)\left(1+y\right)y}}{1+\left(\sqrt{2}+1\right)y+\sqrt{2\left(\sqrt{2}+1\right)\left(1+y\right)y}}\right|
+\frac{\ln\left(\sqrt{2}+1+\sqrt{2\left(\sqrt{2}+1\right)}\,\right)}{\sqrt{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
发表于 2019-2-8 22:21:37 | 显示全部楼层
应楼主要求。回复如下
\[\color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}{\rm\,d}x\!{\rm\,d}y=4\sqrt{\frac{2}{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)}\]
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 2019-2-9 10:24:39 | 显示全部楼层
\[\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}{\rm\,d}x\!{\rm\,d}y = 2\int _0^1dx\int _0^x\frac{\sqrt{x+y}}{x^2+y^2}dy\]
\[ = 2\int_0^1(-\frac{\log \left(\left(\sqrt{2}+1\right) x+2 \sqrt{\sqrt{2}+1} x+x\right)}{2 \sqrt{2 \left(\sqrt{2}+1\right)} \sqrt{x}}+\frac{\log \left(\left(\sqrt{2}+1\right) x-2 \sqrt{\sqrt{2}+1} x+x\right)}{2 \sqrt{2 \left(\sqrt{2}+1\right)} \sqrt{x}}+\frac{\log \left(\left(\sqrt{2}+1\right) x+\sqrt{2 \left(\sqrt{2}+1\right)} x\right)}{2 \sqrt{2 \left(\sqrt{2}+1\right)} \sqrt{x}}-\frac{\log \left(\left(\sqrt{2}+1\right) x-\sqrt{2 \left(\sqrt{2}+1\right)} x\right)}{2 \sqrt{2 \left(\sqrt{2}+1\right)} \sqrt{x}}+\frac{\sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)} \tan ^{-1}\left(\frac{\sqrt{\sqrt{2}+1}+2}{\sqrt{\sqrt{2}-1}}\right)}{\sqrt{x}}+\frac{\sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}+1\right)}-2}{\sqrt{2 \left(\sqrt{2}-1\right)}}\right)}{\sqrt{x}}-\frac{\sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)} \tan ^{-1}\left(\frac{\sqrt{\sqrt{2}+1}-2}{\sqrt{\sqrt{2}-1}}\right)}{\sqrt{x}}-\frac{\sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}+1\right)}+2}{\sqrt{2 \left(\sqrt{2}-1\right)}}\right)}{\sqrt{x}})dx\]
\[=2 \sqrt{2 \left(\sqrt{2}+1\right)} \tan ^{-1}\left(\frac{2}{7} \sqrt{10 \sqrt{2}+2}\right)\]
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
 楼主| 发表于 2019-2-11 15:09:57 | 显示全部楼层
本帖最后由 葡萄糖 于 2019-2-11 15:11 编辑
wayne 发表于 2019-2-8 22:21
应楼主要求。回复如下
\[\color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y=4\sqrt{\frac{2}{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)}\]...


\begin{align*}   
\color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y}   
&=2\int_0^1\!\int_{\sqrt{v}}^{\sqrt{1+v}}\frac{u^2}{\left(u^2-v\right)^2+v^2}\mathrm{d}u\mathrm{d}v\\   
&=2\int_0^1\!\int_0^{u^2}\frac{u^2}{\left(u^2-v\right)^2+v^2}\mathrm{d}v\mathrm{d}u+2\int_1^{\sqrt{2}}\!\int_{u^2-1}^1\frac{u^2}{\left(u^2-v\right)^2+v^2}\mathrm{d}v\mathrm{d}u\\   
&=\pi+4\int_1^{\sqrt{2}}\arctan\left(\frac{2}{u^2}-1\right)\mathrm{d}u\\   
&=8\int_1^{\sqrt{2}}\frac{u^2}{\left(u^2-1\right)^2+1}\mathrm{d}u\\   
&=4\int_1^{\sqrt{2}}\frac{\left(u^2+\sqrt{2}\,\right)+\left(u^2-\sqrt{2}\,\right)}{u^4+2-2u^2}\mathrm{d}u\\   
&=4\int_1^{\sqrt{2}}\left(\frac{1+\frac{\sqrt{2}\,}{u^2}}{u^2+\frac{2}{u^2}-2}\right)\mathrm{d}u   
+4\int_1^{\sqrt{2}}\left(\frac{1-\frac{\sqrt{2}\,}{u^2}}{u^2+\frac{2}{u^2}-2}\right)\mathrm{d}u\\   
&=4\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(u-\frac{\sqrt{2}\,}{u}\right)}{\left(u-\frac{\sqrt{2}\,}{u}\right)^2+2\left(\sqrt2-1\right)}   
+4\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(u+\frac{\sqrt{2}\,}{u}\right)}{\left(u+\frac{\sqrt{2}\,}{u}\right)^2-2\left(\sqrt2+1\right)}\\   
&=4\sqrt{\frac{2}{\sqrt2-1}\,}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)+0 \\  
\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
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