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标题: 二次无理分式的二重瑕积分 [打印本页]

作者: 青青子衿    时间: 2019-1-8 11:06     标题: 二次无理分式的二重瑕积分

本帖最后由 青青子衿 于 2019-2-4 12:46 编辑

\[ \color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}{\rm\,d}x\!{\rm\,d}y} \]
作者: 青青子衿    时间: 2019-1-8 11:07

本帖最后由 青青子衿 于 2019-2-4 13:00 编辑

回复 1# 青青子衿
\begin{align*}   
&&\int_0^1\frac{\sqrt{x+y\,}}{x^2+y^2}\mathrm{d}x&
=\int_{\sqrt{\,y\,}\,}^{\sqrt{\,1+y\,}}\frac{u}{\left(u^2-y\right)^2+y^2}\left(2u\right)\mathrm{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*}
作者: 青青子衿    时间: 2019-1-9 17:24

本帖最后由 青青子衿 于 2019-2-7 14:04 编辑

回复 2# 青青子衿
回复  青青子衿
\begin{align*}
\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)}
\end{align*}...
青青子衿 发表于 2019-1-8 11:07

\begin{align*}   
&&&\color{red}{\int_0^1\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)\!{\rm\,d}y}\\  
&&=&\,\frac{\color{orange}{2}}{\sqrt{2\left(\sqrt{2}-1\right)}}\int_0^1\arctan\left(\frac{\sqrt{\frac{1+t^2}{t^2}}-\sqrt{\frac{2t^2}{1+t^2}}}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)\!{\rm\,d}t\\  
&&=&\,\left.\frac{\color{orange}{2}t}{\sqrt{2\left(\sqrt{2}-1\right)}}\arctan\left(\frac{\sqrt{\frac{1+t^2}{t^2}}-\sqrt{\frac{2t^2}{1+t^2}}}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)\right|_0^1
{\color{violet}{+}}\color{orange}{2}\int_0^1\left(\frac{1+\left(\sqrt{2}+1\right)\!t^2}{\big(1+t^4\big)\sqrt{1+t^2}}\,t\right)\!{\rm\,d}t\\
&&=&\,\frac{\color{orange}{2}}{\sqrt{2\left(\sqrt{2}-1\right)}}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\,\right)
{\color{violet}{+}}\int_0^1\frac{1+\left(\sqrt{2}+1\right)\!y}{\big(1+y^2\big)\sqrt{1+y}\,}\!{\rm\,d}y\\
&&=\,&\boxed{\Big(1+\sqrt2\Big)\sqrt{\frac{2}{\sqrt{2}-1}}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\,\right)}
\end{align*}

\begin{align*}
&&&\frac{\rm\,d}{{\rm\,d}t}\arctan\left(\frac{\sqrt{\frac{1+t^2}{t^2}}-\sqrt{\frac{2t^2}{1+t^2}}}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)\\
&&=\,&\dfrac{-1}{1+\frac{1+\sqrt{2}}{2}\left(\sqrt{\frac{1+t^2}{t^2}}-\sqrt{\frac{2t^2}{1+t^2}}\,\right)^2}\cdot\dfrac{\sqrt{\frac{1+\sqrt{2}}{2}}\left(\sqrt{\frac{1+t^2}{t^2}}+\sqrt{\frac{2t^2}{1+t^2}}\,\right)}{t\left(1+t^2\right)}\\
&&=\,&-\dfrac{\sqrt{2\left(\sqrt{2}-1\right)}\,t}{1+t^4}\left(\sqrt{\frac{1+t^2}{t^2}}+\sqrt{\frac{2t^2}{1+t^2}}\,\right)\\
&&=\,&-\sqrt{2\left(\sqrt{2}-1\right)}\left(\frac{\sqrt{1+t^2}}{1+t^4}+\frac{\sqrt{2}\,t^2}{\big(1+t^4\big)\sqrt{1+t^2}}\,\right)\\
&&=\,&-\sqrt{2\left(\sqrt{2}-1\right)}\left(\frac{1+\left(\sqrt{2}+1\right)\!t^2}{\big(1+t^4\big)\sqrt{1+t^2}}\,\right)\\
\end{align*}

\begin{align*}  
&&&\int_0^1\frac{1+\left(\sqrt{2}+1\right)\!y}{\big(1+y^2\big)\sqrt{1+y}\,}\!{\rm\,d}y\\  
&&=\,&2\int_1^{\sqrt2}\frac{1+\left(\sqrt{2}+1\right)\!\left(v^2-1\right)}{1+\left(v^2-1\right)^2\,}\!{\rm\,d}v\\  
&&=\,&2\int_1^{\sqrt{2}}\frac{\left(\sqrt{2}+1\right)v^2}{1+\left(v^2-1\right)^2}\!{\rm\,d}v-2\int_1^{\sqrt{2}}\frac{\sqrt{2}}{1+\left(v^2-1\right)^2}\!{\rm\,d}v\\
&&=\,&\color{red}{\frac{2}{\sqrt{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)}
\end{align*}

\begin{align*}
&&&2\int_1^{\sqrt{2}}\frac{v^2}{1+\left(v^2-1\right)^2}\!{\rm\,d}v\\

&&=\,&\int_1^{\sqrt{2}}\frac{\left(v^2+\sqrt{2}\,\right)+\left(v^2-\sqrt{2}\,\right)}{v^4+2-2v^2}{\rm\,d}v\\
&&=\,&\int_1^{\sqrt{2}}\left(\frac{1+\frac{\sqrt{2}\,}{v^2}}{v^2+\frac{2}{v^2}-2}\right){\rm\,d}v
+\int_1^{\sqrt{2}}\left(\frac{1-\frac{\sqrt{2}\,}{v^2}}{v^2+\frac{2}{v^2}-2}\right){\rm\,d}v\\
&&=\,&\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(v-\frac{\sqrt{2}\,}{v}\right)}{\left(v-\frac{\sqrt{2}\,}{v}\right)^2+2\left(\sqrt2-1\right)}
+\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(v+\frac{\sqrt{2}\,}{v}\right)}{\left(v+\frac{\sqrt{2}\,}{v}\right)^2-2\left(\sqrt2+1\right)}\\
&&=\,&\sqrt{\frac{2}{\sqrt2-1}\,}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)+0
\end{align*}

\begin{align*}   
&&&2\int_1^{\sqrt{2}}\frac{1}{1+\left(v^2-1\right)^2}\!{\rm\,d}v\\   
&&=\,&\frac{1}{\sqrt2\,}\int_1^{\sqrt{2}}\frac{\left(\sqrt2+v^2\right)+\left(\sqrt2-v^2\right)}{v^4+2-2v^2}{\rm\,d}v\\   
&&=\,&\frac{1}{\sqrt2\,}\int_1^{\sqrt{2}}\left(\frac{\frac{\sqrt{2}}{v^2}+1}{v^2+\frac{2}{v^2}-2}\right){\rm\,d}v   
+\frac{1}{\sqrt2\,}\int_1^{\sqrt{2}}\left(\frac{\frac{\sqrt{2}}{v^2}-1}{v^2+\frac{2}{v^2}-2}\right){\rm\,d}v\\   
&&=\,&\frac{1}{\sqrt2\,}\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(v-\frac{\sqrt{2}\,}{v}\right)}{\left(v-\frac{\sqrt{2}\,}{v}\right)^2+2\left(\sqrt2-1\right)}   
\color{red}{-}\frac{1}{\sqrt2\,}\int_1^{\sqrt{2}}\frac{{\rm\,d}\left(v+\frac{\sqrt{2}\,}{v}\right)}{\left(v+\frac{\sqrt{2}\,}{v}\right)^2-2\left(\sqrt2+1\right)}\\   
&&=\,&\frac{1}{\sqrt{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)+0  
\end{align*}
作者: 青青子衿    时间: 2019-1-11 12:25

本帖最后由 青青子衿 于 2019-1-11 19:49 编辑

回复 2# 青青子衿
回复  青青子衿
\begin{align*}
\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|
}
\end{align*}
...
青青子衿 发表于 2019-1-8 11:07

\begin{align*}      
&&&\color{red}{\int_0^1\frac{1}{2\sqrt{2\left(\sqrt{2}\,{\color{gray}{+}}\,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|\!{\rm\,d}y}\\     
&&=&\,\frac{\color{orange}{1}}{\sqrt{2\left(\sqrt{2}+1\right)}}\int_0^1\ln\left|\frac{1+\left(\sqrt{2}+1\right)t^2-t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}{1+\left(\sqrt{2}+1\right)t^2+t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}\right|\!{\rm\,d}t\\     
&&=&\,\left.\frac{t}{\sqrt{2\left(\sqrt{2}+1\right)}}\ln\left|\frac{1+\left(\sqrt{2}+1\right)t^2-t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}{1+\left(\sqrt{2}+1\right)t^2+t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}\right|\right|_0^1   
{\color{violet}{+}}\color{orange}{2}\int_0^1\left(\frac{1-\left(\sqrt{2}-1\right)\!t^2}{\big(1+t^4\big)\sqrt{1+t^2}}\,t\right)\!{\rm\,d}t\\   
&&=&\,\frac{2\,\ln\left(\sqrt{2}+1-\sqrt{2\left(\sqrt{2}+1\right)}\,\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}  
{\color{violet}{+}}\int_0^1\frac{1-\left(\sqrt{2}-1\right)\!y}{\big(1+y^2\big)\sqrt{1+y}\,}\!{\rm\,d}y\\
&&=\,&\boxed{\frac{2\,\ln\left(\sqrt{2}+1-\sqrt{2\left(\sqrt{2}+1\right)}\,\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+\Big(2-\sqrt2\Big)\sqrt{\frac{2}{\sqrt{2}-1}}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\,\right)}   
\end{align*}

\begin{align*}
&&&\frac{\rm\,d}{{\rm\,d}t}\ln\left|\frac{1+\left(\sqrt{2}+1\right)t^2-t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}{1+\left(\sqrt{2}+1\right)t^2+t\sqrt{2\left(\sqrt{2}+1\right)\left(1+t^2\right)}}\right|\\

&&=\,&-2\sqrt{2\left(\sqrt{2}+1\right)}\left(\frac{1-\left(\sqrt{2}-1\right)\!t^2}{\big(1+t^4\big)\sqrt{1+t^2}}\,\right)\\
\end{align*}

\begin{align*}   
&&&\int_0^1\frac{1-\left(\sqrt{2}-1\right)\!y}{\big(1+y^2\big)\sqrt{1+y}\,}\!{\rm\,d}y\\   
&&=\,&2\int_1^{\sqrt2}\frac{1-\left(\sqrt{2}-1\right)\!\left(v^2-1\right)}{1+\left(v^2-1\right)^2\,}\!{\rm\,d}v\\   
&&=\,&-2\int_1^{\sqrt{2}}\frac{\left(\sqrt{2}-1\right)v^2}{1+\left(v^2-1\right)^2}\!{\rm\,d}v+2\int_1^{\sqrt{2}}\frac{\sqrt{2}}{1+\left(v^2-1\right)^2}\!{\rm\,d}v\\  
&&=\,&\color{red}{\big(2-\sqrt2\,\big)\sqrt{\frac{2}{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)}  
\end{align*}
作者: 青青子衿    时间: 2019-1-11 20:17

本帖最后由 青青子衿 于 2019-2-2 16:16 编辑
\[ \color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}{\rm\,d}x\!{\rm\,d}y} \]
青青子衿 发表于 2019-1-8 11:06

\[\color{black}{\int_0^1\!\int_0^1\frac{\sqrt{x+y\,}}{\,\,x^2+y^2}{\rm\,d}x\!{\rm\,d}y=4\sqrt{\frac{2}{\sqrt2-1}}\arctan\left(\sqrt{\frac{\sqrt2-1}{2}\,}\,\right)}\]
作者: zhcosin    时间: 2019-1-25 08:56

膜,还是高数有意思些。
作者: 青青子衿    时间: 2019-2-11 12:50

本帖最后由 青青子衿 于 2019-2-11 15:08 编辑

\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*}

\begin{align*}
\color{black}{\int_0^1\!\int_0^1\frac{\left(x+y\right)\sqrt{x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y}
&=4\int_0^1\!\int_{\sqrt{v}}^{\sqrt{1+v}}\frac{u^2\left(u^2-v\right)}{\left(u^2-v\right)^2+v^2}\mathrm{d}u\mathrm{d}v\\
&=4\int_0^1\!\int_0^{u^2}\frac{u^2\left(u^2-v\right)}{\left(u^2-v\right)^2+v^2}\mathrm{d}v\mathrm{d}u+4\int_1^{\sqrt{2}}\!\int_{u^2-1}^1\frac{u^2\left(u^2-v\right)}{\left(u^2-v\right)^2+v^2}\mathrm{d}v\mathrm{d}u\\
&=\frac{\pi}{3}+4\int_1^{\sqrt{2}}u^2\arctan\left(\frac{2}{u^2}-1\right)\mathrm{d}u\\
&=\frac{8}{3}\int_1^{\sqrt{2}}\frac{u^4}{\left(u^2-1\right)^2+1}\mathrm{d}u\\
&=\frac{8\left(\sqrt{2}-1\right)}{3}+\frac{16}{3}\int_1^{\sqrt{2}}\frac{u^2-1}{\left(u^2-1\right)^2+1}\mathrm{d}u\\
&=\frac{8\left(\sqrt{2}-1\right)}{3}+\frac{8\sqrt{\sqrt{2}-1}}{3}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\,\right)
\end{align*}

\begin{align*}
\color{black}{\int_0^1\!\int_0^1\frac{\left(x+y\right)\sqrt{1+x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y}
&=4\int_0^1\!\int_{\sqrt{1+v}}^{\sqrt{2+v}}\frac{u^2\left(u^2-v-1\right)}{\left(u^2-v-1\right)^2+v^2}\mathrm{d}u\mathrm{d}v\\
&=4\int_1^{\sqrt{2}}\!\int_0^{u^2-1}\frac{u^2\left(u^2-v-1\right)}{\left(u^2-v-1\right)^2+v^2}\mathrm{d}v\mathrm{d}u+4\int_{\sqrt{2}}^{\sqrt{3}}\!\int_{u^2-2}^1\frac{u^2\left(u^2-v-1\right)}{\left(u^2-v-1\right)^2+v^2}\mathrm{d}v\mathrm{d}u\\
&=\frac{\left(2\sqrt2-1\right)\pi}{3}+4\int_{\sqrt{2}}^{\sqrt{3}}u^2\arctan\left(\frac{2}{u^2-1}-1\right)\mathrm{d}u\\
&=-\frac{\pi}{3}+\frac{8}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{u^4}{\left(u^2-2\right)^2+1}\mathrm{d}u\\
&=-\frac{\pi}{3}+\frac{8\left(\sqrt{3}-\sqrt{2}\,\right)}{3}+\frac{8}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{4u^2-5}{\left(u^2-2\right)^2+1}\mathrm{d}u\\
\end{align*}

\begin{align*}
&\frac{8}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{4u^2-5}{\left(u^2-2\right)^2+1}\mathrm{d}u\\
=&\frac{4}{3}\sqrt{\frac{5\sqrt{5}+2}{2}}\arctan\left(\frac{1}{2}\sqrt{\frac{5\sqrt{5}+4\sqrt{6}+1}{5}}\,\right)+\frac{4}{3}\sqrt{\frac{5\sqrt{5}-2}{2}}\operatorname{arctanh}\left(\frac{1}{2}\sqrt{\frac{5\sqrt{5}-4\sqrt{6}-1}{5}}\,\right)
\end{align*}




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