Coxeter积分计算
Posted eufisky
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Coxeter积分计算相关的知识,希望对你有一定的参考价值。
egin{align*}
&int_0^{frac{pi}{3}}{arccos left( frac{1-cos x}{ ext{2}cos x}
ight) dx}=int_0^{frac{pi}{3}}{ ext{2}arctan sqrt{frac{ ext{3}cos x-1}{cos x+1}}dx}
\
&=int_0^{pi}{ ext{4}arctan sqrt{frac{ ext{3}cos 2y-1}{cos 2y+1}}dy}quad left( x=2y
ight)
\
&=int_0^{frac{pi}{6}}{ ext{4}arctan left( frac{sqrt{1- ext{3}sin ^2y}}{cos y}
ight) dy}=int_0^{frac{pi}{6}}{4left[ frac{pi}{2}-arctan left( frac{cos y}{sqrt{1- ext{3}sin ^2y}}
ight)
ight] dy}
\
&=frac{pi ^2}{3}-4int_0^{frac{pi}{6}}{arctan left( frac{cos y}{sqrt{1- ext{3}sin ^2y}}
ight) dy}
\
&=frac{pi ^2}{3}-4int_0^{frac{pi}{6}}{int_0^1{frac{cos y}{sqrt{1- ext{3}sin ^2y}}frac{dt}{1-left( frac{1-sin ^2y}{1- ext{3}sin ^2y}
ight) t^2}dy}}
\
&=frac{pi ^2}{3}-int_0^{frac{pi}{6}}{int_0^1{frac{ ext{4}cos ysqrt{1- ext{3}sin ^2y}dt}{left( 1- ext{3}sin ^2y
ight) +left( 1-sin ^2y
ight) t^2}dy}}
\
&=frac{pi ^2}{3}-int_0^{frac{pi}{3}}{int_0^1{frac{4sqrt{3}cos ^2wdt}{ ext{3}cos ^2w+left( 2+cos ^2w
ight) t^2}dw}}quad left( sin w=sqrt{3}sin y
ight)
\
&=frac{pi ^2}{3}-int_0^{frac{pi}{3}}{int_0^1{frac{4sqrt{3}sec ^2wdt}{left[ left( 3+3t^2
ight) +2t^2 an ^2w
ight] left( 1+ an ^2w
ight)}dw}}
\
&=frac{pi ^2}{3}-int_0^{sqrt{3}}{int_0^1{frac{4sqrt{3}dtds}{left[ left( 3+3t^2
ight) +2t^2s^2
ight] left( 1+s^2
ight)}}} left( s= an w
ight)
\
&=frac{pi ^2}{3}-int_0^{sqrt{3}}{int_0^1{frac{4sqrt{3}}{t^2+3}left( frac{1}{1+s^2}-frac{2t^2}{left( 3t^2+3
ight) +2t^2s^2}
ight) dtds}}
\
&=frac{pi ^2}{3}-int_0^1{frac{4sqrt{3}}{t^2+3}left[ frac{pi}{3}-sqrt{frac{2t^2}{3t^2+3}}arctan left( sqrt{frac{2t^2}{t^2+1}}
ight)
ight] dt}
\
&=frac{pi ^2}{9}+4sqrt{2}int_0^1{frac{t}{left( t^2+3
ight) sqrt{t^2+1}}arctan left( frac{tsqrt{2}}{sqrt{t^2+1}}
ight) dt}
\
&=frac{pi ^2}{9}+left[ ext{4} an ^{-1}left( frac{sqrt{t^2+1}}{sqrt{2}}
ight) an ^{-1}left( frac{tsqrt{2}}{sqrt{t^2+1}}
ight)
ight] _{0}^{1}-4sqrt{2}int_0^1{frac{1}{left( 3t^2+1
ight) sqrt{t^2+1}}} an ^{-1}left( frac{sqrt{t^2+1}}{sqrt{2}}
ight) dt
\
&=frac{13pi ^2}{36}-4sqrt{2}int_0^1{frac{1}{left( 3t^2+1
ight) sqrt{t^2+1}} an ^{-1}left( frac{sqrt{t^2+1}}{sqrt{2}}
ight) dt}
\
&=frac{5pi ^2}{36}-int_0^1{frac{4}{3t^2+1}int_0^1{frac{1}{1+left( frac{t^2+1}{2}
ight) u^2}}dudt}
\
&=frac{13pi ^2}{36}-4int_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}}
ight] dudt}}
\
&=frac{5pi ^2}{36}+4int_0^1{frac{u}{left( u^2+3
ight) sqrt{u^2+2}} an ^{-1}left( frac{u}{sqrt{u^2+2}}
ight) du}
\
&=frac{5pi ^2}{36}+4left[ an ^{-1}sqrt{u^2+2} an ^{-1}left( frac{u}{sqrt{u^2+2}}
ight)
ight] _{0}^{1}-4int_0^1{frac{ an ^{-1}sqrt{u^2+2}}{left( u^2+1
ight) sqrt{u^2+2}}du}
\
&=frac{13pi ^2}{36}-4int_0^1{frac{ an ^{-1}sqrt{u^2+2}}{left( u^2+1
ight) sqrt{u^2+2}}du}=frac{13pi ^2}{36}-frac{5pi ^2}{24}=frac{11pi ^2}{72}.
end{align*}
以上是关于Coxeter积分计算的主要内容,如果未能解决你的问题,请参考以下文章