Coxeter积分计算

标签：udt   begin   text   amp   计算   cos   ext   ali   dtd   

\begin{align*}
&\int_0^{\frac{\pi}{3}}{\arccos \left( \frac{1-\cos x}{\text{2}\cos x} \right) dx}=\int_0^{\frac{\pi}{3}}{\text{2}\arctan \sqrt{\frac{\text{3}\cos x-1}{\cos x+1}}dx}
\\
&=\int_0^{\pi}{\text{4}\arctan \sqrt{\frac{\text{3}\cos 2y-1}{\cos 2y+1}}dy}\quad \left( x=2y \right)
\\
&=\int_0^{\frac{\pi}{6}}{\text{4}\arctan \left( \frac{\sqrt{1-\text{3}\sin ^2y}}{\cos y} \right) dy}=\int_0^{\frac{\pi}{6}}{4\left[ \frac{\pi}{2}-\arctan \left( \frac{\cos y}{\sqrt{1-\text{3}\sin ^2y}} \right) \right] dy}
\\
&=\frac{\pi ^2}{3}-4\int_0^{\frac{\pi}{6}}{\arctan \left( \frac{\cos y}{\sqrt{1-\text{3}\sin ^2y}} \right) dy}
\\
&=\frac{\pi ^2}{3}-4\int_0^{\frac{\pi}{6}}{\int_0^1{\frac{\cos y}{\sqrt{1-\text{3}\sin ^2y}}\frac{dt}{1-\left( \frac{1-\sin ^2y}{1-\text{3}\sin ^2y} \right) t^2}dy}}
\\
&=\frac{\pi ^2}{3}-\int_0^{\frac{\pi}{6}}{\int_0^1{\frac{\text{4}\cos y\sqrt{1-\text{3}\sin ^2y}dt}{\left( 1-\text{3}\sin ^2y \right) +\left( 1-\sin ^2y \right) t^2}dy}}
\\
&=\frac{\pi ^2}{3}-\int_0^{\frac{\pi}{3}}{\int_0^1{\frac{4\sqrt{3}\cos ^2wdt}{\text{3}\cos ^2w+\left( 2+\cos ^2w \right) t^2}dw}}\quad \left( \sin w=\sqrt{3}\sin y \right)
\\
&=\frac{\pi ^2}{3}-\int_0^{\frac{\pi}{3}}{\int_0^1{\frac{4\sqrt{3}\sec ^2wdt}{\left[ \left( 3+3t^2 \right) +2t^2\tan ^2w \right] \left( 1+\tan ^2w \right)}dw}}
\\
&=\frac{\pi ^2}{3}-\int_0^{\sqrt{3}}{\int_0^1{\frac{4\sqrt{3}dtds}{\left[ \left( 3+3t^2 \right) +2t^2s^2 \right] \left( 1+s^2 \right)}}}\ \ \left( s=\tan w \right)
\\
&=\frac{\pi ^2}{3}-\int_0^{\sqrt{3}}{\int_0^1{\frac{4\sqrt{3}}{t^2+3}\left( \frac{1}{1+s^2}-\frac{2t^2}{\left( 3t^2+3 \right) +2t^2s^2} \right) dtds}}
\\
&=\frac{\pi ^2}{3}-\int_0^1{\frac{4\sqrt{3}}{t^2+3}\left[ \frac{\pi}{3}-\sqrt{\frac{2t^2}{3t^2+3}}\arctan \left( \sqrt{\frac{2t^2}{t^2+1}} \right) \right] dt}
\\
&=\frac{\pi ^2}{9}+4\sqrt{2}\int_0^1{\frac{t}{\left( t^2+3 \right) \sqrt{t^2+1}}\arctan \left( \frac{t\sqrt{2}}{\sqrt{t^2+1}} \right) dt}
\\
&=\frac{\pi ^2}{9}+\left[ \text{4}\tan ^{-1}\left( \frac{\sqrt{t^2+1}}{\sqrt{2}} \right) \tan ^{-1}\left( \frac{t\sqrt{2}}{\sqrt{t^2+1}} \right) \right] _{0}^{1}-4\sqrt{2}\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}}\tan ^{-1}\left( \frac{\sqrt{t^2+1}}{\sqrt{2}} \right) dt
\\
&=\frac{13\pi ^2}{36}-4\sqrt{2}\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}\tan ^{-1}\left( \frac{\sqrt{t^2+1}}{\sqrt{2}} \right) dt}
\\
&=\frac{5\pi ^2}{36}-\int_0^1{\frac{4}{3t^2+1}\int_0^1{\frac{1}{1+\left( \frac{t^2+1}{2} \right) u^2}}dudt}
\\
&=\frac{13\pi ^2}{36}-4\int_0^1{\int_0^1{\frac{1}{u^2+3}\left[ \frac{1}{t^2+\frac{1}{3}}-\frac{1}{t^2+\frac{u^2+2}{u^2}} \right] dudt}}
\\
&=\frac{5\pi ^2}{36}+4\int_0^1{\frac{u}{\left( u^2+3 \right) \sqrt{u^2+2}}\tan ^{-1}\left( \frac{u}{\sqrt{u^2+2}} \right) du}
\\
&=\frac{5\pi ^2}{36}+4\left[ \tan ^{-1}\sqrt{u^2+2}\tan ^{-1}\left( \frac{u}{\sqrt{u^2+2}} \right) \right] _{0}^{1}-4\int_0^1{\frac{\tan ^{-1}\sqrt{u^2+2}}{\left( u^2+1 \right) \sqrt{u^2+2}}du}
\\
&=\frac{13\pi ^2}{36}-4\int_0^1{\frac{\tan ^{-1}\sqrt{u^2+2}}{\left( u^2+1 \right) \sqrt{u^2+2}}du}=\frac{13\pi ^2}{36}-\frac{5\pi ^2}{24}=\frac{11\pi ^2}{72}.
\end{align*}

Coxeter积分计算

标签：udt   begin   text   amp   计算   cos   ext   ali   dtd   

原文地址：https://www.cnblogs.com/Eufisky/p/10063602.html