一道有关对数函数的定积分

青青子衿

\[ \int_0^{y_0}\ln\left(1-x-y+\sqrt{x^2+y^2+\left(1-x-y\right)^2}\,\right){\rm\,d}y \]

青青子衿

\begin{align*}
&&&\dfrac{\rm\,d}{{\rm\,d}y}\ln\left(1-x-y+\sqrt{x^2+y^2+\left(1-x-y\right)^2}\,\right)\\
&&=&\,\dfrac{-1-\frac{1-x-2y}{\sqrt{x^2+y^2+\left(1-x-y\right)^2}}}{1-x-y+\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\\
&&=&\,-\,\frac{\sqrt{x^2+y^2+\left(1-x-y\right)^2}+\left(1-x-2y\right)}{\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\cdot\dfrac{\sqrt{x^2+y^2+\left(1-x-y\right)^2}-\left(1-x-y\right)}{x^2+y^2+\left(1-x-y\right)^2-\left(1-x-y\right)^2}\\
&&=&\,\frac{y-\sqrt{x^2+y^2+\left(1-x-y\right)^2}-\left(1-x-y\right)}{\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\cdot\dfrac{\sqrt{x^2+y^2+\left(1-x-y\right)^2}-\left(1-x-y\right)}{x^2+y^2}\\
&&=&\,\frac{y\sqrt{x^2+y^2+\left(1-x-y\right)^2}-\left(y-xy-y^2\right)-\left(x^2+y^2\right)}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\\
&&=&\,\frac{y\sqrt{x^2+y^2+\left(1-x-y\right)^2}-y+xy-x^2}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\\
&&=&\,\frac{y}{x^2+y^2}-\,\frac{\left(1-x\right)y}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}-\frac{x^2}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\\
&&=&\,\frac{y}{x^2+y^2}-\,\frac{x^2+\left(1-x\right)y}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}\\
\end{align*}

\begin{align*}
&&&\int_0^{y_0}\frac{y^2}{x^2+y^2}{\rm\,d}y\\
&&=&\int_0^{y_0}\frac{x^2+y^2}{x^2+y^2}{\rm\,d}y-\int_0^{y_0}\frac{x^2}{x^2+y^2}{\rm\,d}y\\
&&=&\,y_{\overset{\,}{0}}-\int_0^{y_0}\frac{x^2}{x^2+y^2}{\rm\,d}y=\,y_{\overset{\,}{0}}-x\int_0^{y_0}\frac{1}{1+\left(\frac{y}{x}\right)^2}{\rm\,d}\left(\frac{y}{x}\right)\\
&&=&\,y_{\overset{\,}{0}}-x\arctan\left(\frac{y_{\overset{\,}{0}}}{x}\right)\\
\end{align*}

\begin{align*}  
&&&\int_0^{y_0}\ln\left(1-x-y+\sqrt{x^2+y^2+\left(1-x-y\right)^2}\,\right){\rm\,d}y\\  
&&=\,\,&\,y_0\ln\left(1-x-y_0+\sqrt{x^2+{y_0}^2+\left(1-x-y_0\right)^2}\,\right)-\int_0^{y_0}\frac{y^2}{x^2+y^2}{\rm\,d}y\\
&&&\,+
\int_0^{y_0}\frac{x^2y+\left(1-x\right)y^2}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}{\rm\,d}y\\
\end{align*}

\begin{align*}
&&&  
\int_0^{y_0}\frac{x^2y+\left(1-x\right)y^2}{\left(x^2+y^2\right)\sqrt{x^2+y^2+\left(1-x-y\right)^2}}{\rm\,d}y\\
&&=&
\int_0^{y_0}\frac{x^2y+\left(1-x\right)y^2}{\left(x^2+y^2\right)\sqrt{x^2+\left(1-x\right)^2+2y^2-2\left(1-x\right)y}}{\rm\,d}y\\
&&=&
\int_0^{y_0}\frac{x^2y+\left(1-x\right)y^2}{\left(x^2+y^2\right)\sqrt{x^2+\frac{\left(1-x\right)^2}{2}+2\left(y-\frac{1-x}{2}\right)^2}}{\rm\,d}y\\
&&=&  
\frac{1}{\sqrt{2\,}}\int_{-\operatorname{arsinh}\left(\frac{q}{p}\right)}^{\operatorname{arsinh}\left(\frac{y_{\overset{\,}{0}}-q}{p}\right)}
\frac{x^2\left(p\sinh\,t+q\right)+\left(1-x\right)\left(p\sinh\,t+q\right)^2}{x^2+\left(p\sinh\,t+q\right)^2}{\rm\,d}t\\

\end{align*}

zhcosin

回复 2# 青青子衿
膜拜青神