Ahmed’s integrals 和 Coxeter’s integrals - Eufisky - The lost book

# Ahmed’s integrals 和 Coxeter’s integrals

Eufisky posted @ 2015年12月11日 21:11 in 数学分析 with tags 积分计算 , 1835 阅读

1. 主要结果

$$A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{p q r}{(r^{2}+1)p^{2}x^{2} + 1} \, dx.$$

\begin{align*} \rho &= \frac{1-pqr}{1+pqr}, & \alpha &= 2\arctan\left( \frac{qr}{\sqrt{r^{2}+1}} \right), \\ \beta &= 2\arctan\left( \frac{rp}{\sqrt{p^{2}+1}} \right), & \gamma &= 2\arctan\left( r\sqrt{q^{2}+1} \right). \end{align*}

\begin{align*} A(p, q, r) &= \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right) \\ &\qquad – \frac{1}{2}\Re \left( \operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho e^{i\alpha}) – \operatorname{Li}_{2}(\rho e^{i\beta}) + \operatorname{Li}_{2}(\rho e^{i\gamma}) \right). \end{align*}

$$A(p, q, r) = A(\rho \mid \alpha, \beta, \gamma)$$

2. 推论

$$A(p, q, r) = \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right).$$

\begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} &= A\left( \frac{1}{\sqrt{2}}, \sqrt{2}, 1 \right) \\ &= A\left( 0 \, \middle| \, \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{96}. \end{align*}

$$A(p, q, r) = \frac{\pi}{4} \left( \alpha + \beta – \gamma \right).$$

$$\Re\operatorname{Li}_{2}(-e^{i\theta}) = \operatorname{Li}_{2}(-1) + \frac{\theta^{2}}{4} \quad |\theta| \leq \pi,$$

\begin{align*} A(-1 \mid \alpha, \beta, \gamma) &= \frac{\pi}{4} \left( \alpha + \beta – \gamma \right). \end{align*}

3. 在Coxeter’s integrals上的应用

\begin{align*} &\int_{0}^{\beta} \arctan \sqrt{\frac{\cos\theta + 1}{a\cos\theta + b}} \, d\theta \\ &= 2 A \left( \sqrt{\frac{a-b}{2} \cdot \frac{1-\cos\beta}{a\cos\beta+ b}}, \sqrt{\frac{2}{a+b}}, \sqrt{\frac{a+b}{a-b}} \right) \\ &= 2 A \left( \frac{1-k}{1+k} \, \middle| \, \alpha, \beta, \gamma\right) \end{align*}

$$k = \left( \frac{1-\cos\beta}{a\cos\beta+ b} \right)^{1/2}, \quad \alpha = \arccos\left(\frac{a-1}{a+1}\right), \quad \gamma = \arccos\left(-\frac{1+b}{1+a}\right).$$定义的.

\begin{align*} \int_{0}^{\frac{\pi}{2}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{2}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( 0 \, \middle| \, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{24}. \end{align*}

\begin{align*} \int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{3}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \, \middle| \, \frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} \right). \end{align*}

$$\operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho^{2}) – \operatorname{Li}_{2}(\rho^{3}) + \frac{1}{3}\operatorname{Li}_{2}(\rho^{6}), \quad \rho = \frac{\sqrt{5}-1}{\sqrt{5}+1}.$$

$$\int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta = \frac{2\pi^{2}}{15}.$$

\begin{align*}\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx}  &= 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{ - 1 + 3\cos x}}{{1 + \cos x}}} dx}  = 2\int_0^{\frac{\pi }{3}} {\left( {\frac{\pi }{2} - \arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} } \right)dx} \\&= \frac{{{\pi ^2}}}{3} - 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} dx}  = \frac{{{\pi ^2}}}{3} - 4A\left( {1\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right).\end{align*}

\begin{align*}A\left( {0\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) + \frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) - \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right)} \right]\\=& \frac{{13{\pi ^2}}}{{288}} .\end{align*}

$\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \int_0^{\frac{\pi }{2}} {\arctan \sqrt {\frac{{\cos x + 1}}{{\cos x}}} dx} = 2A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right)$

\begin{align*}A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) + \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) - \frac{{2\pi }}{3}\left( {2\pi  - \frac{{2\pi }}{3}} \right)} \right]\\&- \frac{1}{2}{\mathop{\rm Re}\nolimits} \left[ { \operatorname{Li}_2\left( { - 1} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) + \operatorname{Li}_2\left( { - {e^{i\frac{{2\pi }}{3}}}} \right)} \right]\\= &\frac{{11{\pi ^2}}}{{144}} - \frac{1}{2}\left[ { - \frac{{{\pi ^2}}}{{12}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{36}}} \right] = \frac{{{\pi ^2}}}{{12}}.\end{align*}

\begin{align*}&{\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right)} \right\} = {\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{ - i\frac{{2\pi }}{3}}}} \right)} \right\} = \frac{1}{2}\left\{ { \operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right) + \operatorname{Li}_2\left( {1 - {e^{i\frac{\pi }{3}}}} \right)} \right\}\\= &\frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \ln {e^{i\frac{\pi }{3}}}\ln {e^{ - i\frac{\pi }{3}}}} \right\} = \frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \left( {i\frac{\pi }{3}} \right)\left( { - i\frac{\pi }{3}} \right)} \right\} = \frac{{{\pi ^2}}}{{36}}.\end{align*}

$\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \frac{{{\pi ^2}}}{6}.$

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2023年9月20日 00:09

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