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  • Coxeter积分计算

    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*}

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