三次无理分式的二重瑕积分

青青子衿

\[ \color{black}{\int_0^1\!\int_0^1\frac{\sqrt[3]{x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y} \]

青青子衿

回复 1# 青青子衿

\begin{align*}      
&&\int_0^1\frac{\sqrt[3]{x+y\,}}{x^2+y^2}\mathrm{d}x&   
=\int_{\sqrt[3]{\,y\,}\,}^{\sqrt[3]{\,1+y\,}}\frac{u}{\left(u^3-y\right)^2+y^2}\left(3u^2\right)\mathrm{d}u\\   
&&&=3\int_{\sqrt[3]{\,y\,}\,}^{\sqrt[3]{\,1+y\,}}\frac{u^3}{\left(u^3-y\right)^2+y^2}\mathrm{d}u\\   
&&&=\frac{3}{2}\int_{\sqrt[3]{\,y\,}\,}^{\sqrt[3]{\,1+y\,}}\frac{\left(u^3+\sqrt[3]{2\,y^2}\,u\,\right)+\left(u^2-\sqrt[3]{2\,y^2}\,u\,\right)}{u^6+2y^2-2u^3y}\mathrm{d}u\\   
&&&=\frac{3}{2}\int_{\sqrt[3]{\,y\,}\,}^{\sqrt[3]{\,1+y\,}}\left(\frac{1+\frac{\sqrt[3]{2\,y^2\,}\,}{u^2}}{u^3+\frac{2\,y^2}{u^3}-2y}\right)\!{\rm\,d}u   
+\frac{3}{2}\int_{\sqrt{\,y\,}\,}^{\sqrt[3]{\,1+y\,}}\left(\frac{1-\frac{\sqrt[3]{2\,y^2}\,}{u^2}}{u^2+\frac{2\,y^2\,}{u^2}-2y}\right)\!{\rm\,d}u\\   
&&&=\\   
  
\end{align*}

青青子衿

\begin{align*}
&&\color{black}{\int_0^1\!\int_0^1\frac{\sqrt[3]{x+y\,}}{\,\,x^2+y^2}\mathrm{d}x\mathrm{d}y}
\quad
&\overset{\Large{???}}{\overline{\overline{\hspace{1cm}}}}\quad18\color{black}{{\large\int}_1^{\sqrt[3]{2\,}}\frac{u^3}{\left(u^3-1\right)^2+1}\mathrm{d}u}\\
\end{align*}
青青子衿 发表于 2019-2-4 12:52

\[ \color{black}{\int_{1}^{\sqrt[3]{\,2\,}}\frac{u^3}{\left(u^3-1\right)^2+1}\mathrm{d}u=I_A+I_B+I_C} \]
\begin{align*}
I_A&=\phantom{+}\frac{\sqrt[6]{\,2\,}}{\,\,18}\sin\left(\frac{\pi}{12}\right)\ln2+\frac{\sqrt[6]{\,2\,}}{\,\,3}\cos\left(\frac{\pi}{12}\right)\arctan\left(\frac{1}{\sqrt{3}\left(1+2\sqrt[3]{\,2\,}\,\right)-2\left(1+\sqrt[3]{\,4\,}\,\right)}\right)\\
I_B&=\phantom{+}\frac{\sqrt[6]{\,2\,}}{\,\,18}\sin\left(\,\frac{\pi}{\,4\,}\,\right)\ln2-\frac{\sqrt[6]{\,2\,}}{3}\cos\left(\,\frac{\pi}{\,4\,}\,\right)\arctan\left(\frac{8\sqrt[3]{\,4\,}-10\sqrt[3]{\,2\,}+1}{23}\right)\\
I_C&=-\frac{\sqrt[6]{\,2\,}}{\,\,18}\cos\left(\frac{\pi}{12}\right)\ln2+\frac{\sqrt[6]{\,2\,}}{\,\,3}\sin\left(\frac{\pi}{12}\right)\arctan\left(\frac{1}{\sqrt{3}\left(1+2\sqrt[3]{\,2\,}\,\right)+2\left(1+\sqrt[3]{\,4\,}\,\right)}\right)
\end{align*}