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有关双十一的一个四重积分

本帖最后由 青青子衿 于 2018-11-12 18:06 编辑

\[\int_0^1\int_0^1\int_0^1\int_0^1\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\kern 2pt}{\rm\,d}x{\rm\,d}y{\rm\,d}z{\rm\,d}w\]
  1. \int_0^1\int_0^1\int_0^1\int_0^1\sqrt{x^2+y^2+z^2+w^2}dxdydzdw
复制代码
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本帖最后由 青青子衿 于 2018-11-15 21:50 编辑

\begin{align*}
\frac{8}{15}+\frac{14\ln3}{15}-\frac{8\sqrt{2}}{15}\arctan\left(\frac{\sqrt{2}}{4}\right)\\
+\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{2}{\sqrt{1+u^2}}+\sqrt{1+\frac{2}{1+u^2}}\right){\rm\,d}u\\
-\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}+\sqrt{1+\frac{1}{1+u^2}}\right){\rm\,d}u\\
-\frac{4}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}\right){\rm\,d}u
\end{align*}

\begin{align*}
&=
\begin{split}
\frac{8}{15}+\frac{14\ln3}{15}-\frac{8\sqrt{2}}{15}\arctan\left(\frac{\sqrt{2}}{4}\right)\\
-\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\sqrt{1+\frac{2}{1+u^2}}\,\right){\rm\,d}u\\
+\frac{2\pi}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3{\rm\,d}u\\
\end{split}\\
&=
\begin{split}
\frac{8}{15}+\frac{14\ln3}{15}-\frac{8\sqrt{2}}{15}\arctan\left(\frac{\sqrt{2}}{4}\right)\\
-\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\sqrt{1+\frac{2}{1+u^2}}\,\right){\rm\,d}u\\
+\frac{7\sqrt{2}\,\pi}{60}+\frac{\pi\ln\left(1+\sqrt{2}\,\right)}{20}\\
\end{split}\\
\end{align*}

\begin{align*}
\arctan\left(\frac{\sqrt{2}}{2}\right)-\arctan\left(\frac{\sqrt{2}}{5}\right)=\arctan\left(\frac{\sqrt{2}}{4}\right)
\end{align*}
\begin{align*}
&\phantom{=}-\arctan\left(\frac{2}{\sqrt{1+x^2}}+\sqrt{1+\frac{2}{1+x^2}}\right)+\arctan\left(\frac{1}{\sqrt{1+x^2}}+\sqrt{1+\frac{1}{1+x^2}}\right)+\frac{1}{2}\arctan\left(\frac{1}{\sqrt{1+x^2}}\right)\\
&=-\arctan\left(\frac{2}{\sqrt{1+x^2}}+\sqrt{1+\frac{2}{1+x^2}}\right)+\frac{1}{2}\arctan\left(\frac{1}{\sqrt{1+x^2}}\right)+\frac{\pi}{4}+\frac{1}{2}\arctan\left(\frac{1}{\sqrt{1+x^2}}\right)\\
&=-\arctan\left(\frac{2}{\sqrt{1+x^2}}+\sqrt{1+\frac{2}{1+x^2}}\right)+\arctan\left(\frac{1}{\sqrt{1+x^2}}\right)+\frac{\pi}{4}\\
&=-\arctan\left(\dfrac{\frac{1}{\sqrt{1+x^2}}+\sqrt{1+\frac{2}{1+x^2}}}{1+\frac{1}{\sqrt{1+x^2}}\left(\frac{2}{\sqrt{1+x^2}}+\sqrt{1+\frac{2}{1+x^2}}\right)}\right)+\frac{\pi}{4}\\
&=-\arctan\left(\dfrac{\sqrt{1+x^2}\left(1+\sqrt{3+x^2}\right)}{3+x^2+\sqrt{3+x^2}}\right)+\frac{\pi}{4}\\
&=-\arctan\left(\sqrt{\dfrac{1+x^2}{3+x^2}}\,\right)+\frac{\pi}{4}\\
&=-\frac{\pi}{2}+\arctan\left(\sqrt{\dfrac{3+x^2}{1+x^2}}\,\right)+\frac{\pi}{4}\\
&=\arctan\left(\sqrt{1+\frac{2}{1+x^2}}\,\right)-\frac{\pi}{4}\\
&=\arctan\left(\frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{\sqrt{3+x^2}+\sqrt{1+x^2}}\,\right)\\
&=\arctan\left(2+x^2-\sqrt{\left(3+x^2\right)\left(1+x^2\right)}\,\right)\\
\end{align*}
...
  1. \frac{8}{15}+\frac{14\ln3}{15}-\frac{8\sqrt{2}}{15}\arctan\left(\frac{\sqrt{2}}{4}\right)
  2. \frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{2}{\sqrt{1+u^2}}+\sqrt{1+\frac{2}{1+u^2}}\right)du-\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}+\sqrt{1+\frac{1}{1+u^2}}\right)du-\frac{4}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}\right)du
  3. -\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\sqrt{1+\frac{2}{1+u^2}}\right)du
  4. \frac{7\sqrt{2}\pi}{60}+\frac{\pi\ln\left(1+\sqrt{2}\right)}{20}
复制代码
...
\begin{align*}
\int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\rm\,d}x{\rm\,d}y=
\begin{array}{r}
\dfrac{ab}{3}\sqrt{a^{\overset{\,}{2}}+b^2+z^2+w^2}
-\dfrac{1}{3}\left(z^2+w^2\right)^{\frac{3}{2}}\arctan\left(\dfrac{ab}{\sqrt{\left(z^2+w^2\right)\left(a^2+b^2+z^2+w^2\right)}}\right)\\
+\dfrac{a}{6}\left(a^2+3z^2+3w^2\right)\ln\left(\dfrac{b}{\sqrt{a^2+z^2+w^2}}+\sqrt{1+\dfrac{b^2}{a^2+z^2+w^2}}\,\right)\\
+\dfrac{b}{6}\left(b^2+3z^2+3w^2\right)\ln\left(\dfrac{a}{\sqrt{b^2+z^2+w^2}}+\sqrt{\dfrac{a^2}{b^2+z^2+w^2}+1}\,\right)\\
\end{array}
\end{align*}

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