Ahmed’s integrals 和 Coxeter’s integrals
最后得到Ahmed’s integrals的一个简单表达式.
1. 主要结果
一般地,我们将含参数$p,q,r$的Ahmed’s integral定义成
$$A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{p q r}{(r^{2}+1)p^{2}x^{2} + 1} \, dx.$$
则下面的定理成立:
定理. 对任意$p, q, r > 0$,记$\rho, \alpha, \beta, \gamma$为
\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*}
我们有
\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*}
基于此定理,只要不产生混乱,我们都可写成
$$A(p, q, r) = A(\rho \mid \alpha, \beta, \gamma)$$
在此,当然, $\rho, \alpha, \beta, \gamma$是出现在上述定理中的参数.
2. 推论
虽然一般的结果包含在dilogarithmic的一支,但他们或者$\rho = 0$或者当$\rho= -1$时,碰巧可以简化.
推论 1.将$\rho, \alpha, \beta, \gamma$记为主定理那样的数.如果$pqr = 1$, 则
$$A(p, q, r) = \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right).$$
在定理中令$\rho = 0$便可得到.
比如, 传统上的Ahmed’s integral可以这样计算:
\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*}
推论 2. 将$\rho, \alpha, \beta, \gamma$记为主定理那样的数. 如果$pqr \to \infty$, 则
$$A(p, q, r) = \frac{\pi}{4} \left( \alpha + \beta – \gamma \right).$$
证.当$pqr \to \infty$, 我们有$\rho \to -1$.由等式
$$\Re\operatorname{Li}_{2}(-e^{i\theta}) = \operatorname{Li}_{2}(-1) + \frac{\theta^{2}}{4} \quad |\theta| \leq \pi,$$
推出
\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
命题. 假设$ a \geq |b|$并且对$\theta \in (0, \beta)$,有$a \cos\theta + b > 0$. 则我们有
\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*}
其中 $k, \alpha, \gamma$是通过
$$k = \left( \frac{1-\cos\beta}{a\cos\beta+ b} \right)^{1/2}, \quad \alpha = \arccos\left(\frac{a-1}{a+1}\right), \quad \gamma = \arccos\left(-\frac{1+b}{1+a}\right).$$定义的.
比如,一个经典的Coxeter’s integrals可以这样计算:
\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*}
许多其它的Coxeter’s integrals可以直接由此命题获得, 结合推论1或推论2. 一些更复杂的情形可能需要polylogarithmic作为阶梯. 例如,
\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*}
为了简化此表达式需要以下的polylogarithmic阶梯工具,
$$\operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho^{2}) – \operatorname{Li}_{2}(\rho^{3}) + \frac{1}{3}\operatorname{Li}_{2}(\rho^{6}), \quad \rho = \frac{\sqrt{5}-1}{\sqrt{5}+1}.$$
由此可知表达式得到的值为 $\pi^{2}/45$, 因此我们最后得到
$$\int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta = \frac{2\pi^{2}}{15}.$$
主页：Sangchul Lee http://www.math.ucla.edu/~sos440/m/index.html
下面给出自己的一些结果:
\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*}
其中用到了\[\cos x = \frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2}}} = \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}},\arctan x + \arctan \frac{1}{x} = \frac{\pi }{2}.\]
而
\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*}
因此\[\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} = \frac{{{11\pi ^2}}}{72}.\]
\[\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \int_0^{\frac{\pi }{2}} {\arctan \sqrt {\frac{{\cos x + 1}}{{\cos x}}} dx} = 2A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right)\]
而
\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*}
这里用到了$\operatorname{Li}_2\left( i \right) = - \frac{{{\pi ^2}}}{{48}} + iC$, 其中$C$是Catalan constant.利用\[\operatorname{Li}_s\left( z \right) = \sum\limits_{k = 1}^\infty {\frac{{{z^k}}}{{{k^s}}}} = z + \frac{{{z^2}}}{{{2^s}}} + \frac{{{z^3}}}{{{3^s}}} + \cdots \]和\[\operatorname{Li}_2\left( x \right) + \operatorname{Li}_2\left( {1 - x} \right) = \frac{1}{6}{\pi ^2} - \ln x\ln \left( {1 - x} \right),\]我们有\[{\mathop{\rm Re}\nolimits} \left\{ { \operatorname{Li}_2\left( z \right)} \right\} = \frac{1}{2}\left\{ { \operatorname{Li}_2\left( z \right) + \operatorname{Li}_2\left( {\bar z} \right)} \right\}\]以及
\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*}
因此
\[\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \frac{{{\pi ^2}}}{6}.\]
2023年9月20日 00:09
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