积分不等式的
双阶乘的估计
双阶乘除式的一个估计
\[\boxed{\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} \sim \sqrt {n\pi } .}\]
证明：
\begin{align*}\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} =\frac{{{{\left[ {\left( {2n} \right)!!} \right]}^2}}}{{\left( {2n} \right)!}} =\frac{{{2^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}} \sim \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \sqrt {n\pi }. \end{align*}
由此可对如下一个关于$\sin x$的整数次幂在$[-\frac{\pi}{2},\frac{\pi}{2}]$上的积分进行数值计算：
\[\int_0^{\frac{\pi }{2}} {{{\sin }^n}xdx} = \left\{ \begin{array}{l}\frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \sim \frac{1}{2}\sqrt{\frac{\pi }{m}} ,n = 2m\\\frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \sim \frac{{\sqrt {m\pi } }}{{2m + 1}},n = 2m + 1\end{array} \right.,m \in {N_ + }.\]
特殊地，我们有：
\[0.250037 \approx I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^{100}}xdx} = 2\int_0^{\frac{\pi }{2}} {{{\sin }^{100}}xdx} = \frac{{99!!}}{{100!!}}\pi \sim \frac{1}{{\sqrt {50\pi } }}\pi = \frac{1}{5}\sqrt {\frac{\pi }{2}} = 0.25003696.\]
由上可知，此误差是很小的。
一个级数求解
\[\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma + \ln 2 + \ln \pi } \right).\]
证明：(Glaisher–Kinkelin constant)\[\boxed{\ln A = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^2}}}{4} - \left( {\frac{{{n^2}}}{2} + \frac{n}{2} + \frac{1}{{12}}} \right)\ln n + \sum\limits_{k = 1}^n {k\ln k} } \right].}\]
(Riemann zeta函数的导函数)\[\boxed{\zeta '\left( s \right) = - \sum\limits_{k = 1}^\infty {\frac{{\ln k}}{{{k^s}}}} .}\]
(Gamma 函数)
\[\Gamma \left( s \right) = \int_0^\infty {{x^{s - 1}}{e^{ - x}}dx}. \]
首先，证明$\zeta'(-1)=\frac{1}{12}-\ln A$.
再证明$\Gamma'(2)=1-\gamma.$
\[\Gamma '\left( 2 \right) = \int_0^\infty {x{e^{ - x}}\ln xdx} = \int_0^\infty {\left( {{e^{ - x}} + {e^{ - x}}\ln x} \right)dx} = 1 - \gamma .\]
利用
\[\zeta \left( s \right) = {2^s}{\pi ^{s - 1}}\sin \frac{{\pi s}}{2}\Gamma \left( {1 - s} \right)\zeta \left( {1 - s} \right).\]
令$s=-1$，我们有$\zeta{-1}=-\frac{1}{12}.$
两边同取对数得
\[\ln \zeta \left( s \right) = s\ln 2 + \left( {s - 1} \right)\ln \pi + \ln \sin \frac{{\pi s}}{2} + \ln \Gamma \left( {1 - s} \right) + \ln \zeta \left( {1 - s} \right).\]
求导，得到\[\frac{{\zeta '\left( s \right)}}{{\zeta \left( s \right)}} = \ln \left( {2\pi } \right) + \frac{\pi }{{2\tan \frac{{\pi s}}{2}}} - \frac{{\Gamma '\left( {1 - s} \right)}}{{\Gamma \left( {1 - s} \right)}} - \frac{{\zeta '\left( {1 - s} \right)}}{{\zeta \left( {1 - s} \right)}}.\]
令$s=-1$，我们有
\[\frac{{\zeta '\left( { - 1} \right)}}{{\zeta \left( { - 1} \right)}} = 12\ln A - 1 = \ln \left( {2\pi } \right) - 1 + \gamma - \frac{{\zeta '\left( 2 \right)}}{{\zeta \left( 2 \right)}}.\]
\[ \Rightarrow \zeta '\left( 2 \right) = \frac{{{\pi ^2}}}{6}\left( {\ln \left( {2\pi } \right) - 12\ln A + \gamma } \right).\]
因此我们得到\[\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma + \ln 2 + \ln \pi } \right).\]
几个欧拉和（Euler sum）的求解
\begin{align}&\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)}; \\&\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)}; \\&\sum\limits_{n = 1}^\infty {\frac{{\left( {\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} } \right)\left( {\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} } \right)}}{{{n^2}}}}; \\&\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}{H_n}}}} ,{H_n} = \sum\limits_{k = 1}^n {\frac{1}{n}}. \end{align}
傅里叶变换求解积分题2
计算
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx.}\]
解：留数理论的一种解答：
注意到
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .\]
若令
\begin{align*}F\left( m \right) &= \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}dx}\\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x + 1} \right)\cos \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x - 1} \right)\cos \left( {mx} \right)}}{{{x^2} - x + 1}}dx}.\end{align*}
\begin{align*} \Rightarrow F'\left( m \right) &= - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} + x} \right)\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} + \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} - x} \right)sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} \\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx}\end{align*}
再令
\[I = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .\]
则
\[I = {\mathop{\rm Im}\nolimits} T.\]
即$\displaystyle T$的虚部为$\displaystyle I$.因此，为了计算积分$\displaystyle I$，只需求出积分
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} \]
即可.先求
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .\]
求得辅助函数
\[\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}\]
在上半平面的奇点只有点$\displaystyle \alpha = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i$（另一个奇点为$\displaystyle \beta = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i$）.于是我们有
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).\]
由于
\[{\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.\]
\[ \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).\]
同理亦得
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).\]
\[ \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}\]
故
\[F'\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) = - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.\]
\[ \Rightarrow F\left( m \right) = - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).\]
\begin{align*} \Rightarrow \int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} &= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}F\left( m \right) \\&= \frac{\pi }{{2\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) = \frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\end{align*}
另解：由Fourier变换公式，我们有
\begin{align*}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) &= \frac{2}{\pi }\int_0^\infty {\cos \left( {mx} \right)dx} \int_0^\infty {{e^{ - \frac{{\sqrt 3 }}{2}u}}\left( {\cos \frac{u}{2} + \sqrt 3 \sin \frac{u}{2}} \right)\cos \left( {ux} \right)du} \\&= \frac{{2\sqrt 3 }}{\pi }\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}.\end{align*}
立得
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\]
2013年武汉大学数学分析考研真题
数学钟的12个问题
上面是自己在网上看到的一个数学钟，设计者真是别有用心，于是利用周末闲暇时光，对上面的12个问题，一一给出说明.
数字1：Euler公式
\[e^{i\pi}+1=0.\]
数字2：
对于
\[\Gamma \left( {x + 1} \right) = \int_0^\infty {{e^{ - v}}{v^x}dv} .\]
我们有
\[\Gamma '\left( {x + 1} \right) = \int_0^\infty {{e^{ - v}}{v^x}\ln vdv} .\]
另外
\[\frac{{\Gamma '\left( {x + 1} \right)}}{{\Gamma \left( {x + 1} \right)}} = - \gamma + \sum\limits_{k = 1}^\infty {\frac{x}{{k\left( {x + k} \right)}}} \Rightarrow \Gamma '\left( 1 \right) = - \gamma .\]
所以我们有
\[\Gamma '\left( 1 \right) = \int_0^\infty {{e^{ - v}}\ln vdv} = - \gamma. \]
由此得
\begin{align*}\int_0^\infty {\frac{{{e^{ - {x^2}}} - {e^{ - x}}}}{x}dx} &= \left[ {\left( {{e^{ - {x^2}}} - {e^{ - x}}} \right)\ln x} \right]_0^\infty + \int_0^\infty {2x{e^{ - {x^2}}}\ln xdx} - \int_0^\infty {{e^{ - x}}\ln xdx} \\&= 0 + \frac{1}{2}\int_0^\infty {{e^{ - t}}{\mathop{\rm lnt}\nolimits} dt} - \int_0^\infty {{e^{ - x}}\ln xdx} = - \frac{1}{2}\int_0^\infty {{e^{ - x}}\ln xdx} = \frac{\gamma }{2}.\end{align*}
数字3：
\[\frac{\pi}{4}=5\arctan{\frac{1}{7}}+2\arctan{\frac{3}{79}}.\]
证明和数字5的大致是一样的，只是计算量偏大了点
\[{\left( {7 - i} \right)^5}\left( {1 + i} \right) = {\left( {48 - 14i} \right)^2}\left( {8 + 6i} \right) = \left( {2108 - 1344i} \right)\left( {8 + 6i} \right) = 8\left( {3116 + 237i} \right) = 4{\left( {79 + 3i} \right)^2}.\]
由此得
\[\arg \left( {1 + i} \right) + \arg {\left( {7 - i} \right)^5} = \arg \left[ {4{{\left( {79 + 3i} \right)}^2}} \right] = \arg {\left( {79 + 3i} \right)^2}.\]
若取$\displaystyle -\pi<\arg\leq\pi$，则由上式可知
\[\frac{\pi }{4} - 5\arctan \frac{1}{7} = 2\arctan \frac{3}{{79}}.\]
即
\[\frac{\pi }{4} =5\arctan \frac{1}{7} + 2\arctan \frac{3}{{79}}.\]
这一结果属于Leonhard Euler.
数字4:
\[\arctan x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}}} .\]
再令$\displaystyle x=1$即可.
数字5:
应用
\[\frac{\pi}{4}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}}.\]
下面是泉叔书上给出的一个较为简便的证明方法
\[{\left( {5 - i} \right)^4}\left( {1 + i} \right) = {\left( {24 - 10i} \right)^2}\left( {1 + i} \right) = \left( {476 - 480i} \right)\left( {1 + i} \right) = 4\left( {239 - i} \right).\]
于是
\[\arg \left( {1 + i} \right) + \arg {\left( {5 - i} \right)^4} = \arg \left[ {4\left( {239 - i} \right)} \right] = \arg \left( {239 - i} \right).\]
若取$\displaystyle -\pi<\arg\leq\pi$，则由上式可知
\[\frac{\pi }{4} - 4\arctan \frac{1}{5} = - \arctan \frac{1}{{239}}.\]
即
\[\frac{\pi }{4} =4\arctan \frac{1}{5} - \arctan \frac{1}{{239}}.\]
这一结果归功于 John Machin.
数字6:
\[\int_0^1 {\frac{{\ln x}}{{x - 1}}dx} = \int_1^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt} = \mathrm{Li}_2\left( 1 \right) = \frac{{{\pi ^2}}}{6}.\]
数字7:
\[\tan \left( {2\arctan \frac{1}{3}} \right) = \frac{{2 \cdot \frac{1}{3}}}{{1 - {{\left( {\frac{1}{3}} \right)}^2}}} = \frac{3}{4}\]
\[\Rightarrow \tan \left( {\frac{\pi }{4} - 2\arctan \frac{1}{3}} \right) = \frac{{1 - \frac{3}{4}}}{{1 + \frac{3}{4}}} = \frac{1}{7}.\]
数字8:
\[\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {2k + 1} \right)}^2}}}} = \sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} - \sum\limits_{k = 1}^\infty {\frac{1}{{{{\left( {2k} \right)}^2}}}} = \frac{3}{4}\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} = \frac{{{\pi ^2}}}{8}.\]
数字9:注意到
\[\prod\limits_{k = 1}^m {\tan \frac{{k\pi }}{{2m + 1}}} = \sqrt {2m + 1}.\]
令$\displaystyle m=4$即可得到.
数字10:
\[\sin \frac{\pi }{{10}} = \frac{{\sqrt 5 - 1}}{4}.\]
数字11：
首先证明：$\displaystyle\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} = \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)}}{3}.$
由Euler公式
\[\left(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}}\right)^{2n}=(-1)^{k}.\]
由牛顿二项式定理
\[\sum_{t=0}^{2n}\binom{2n}{t}\left(\cos{\frac{k\pi}{2n}}\right)^t \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-t}=(-1)^{k}.\]
只考虑虚部
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(\cos{\frac{k\pi}{2n}}\right)^{2r+1} \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.\]
两边同除 $\displaystyle \left(\cos{\frac{k\pi}{2n}}\right)^{2n}$
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.\]
两边同乘$\displaystyle i\tan{\frac{k\pi}{2n}}$
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r}=0.\]
故 $\displaystyle \tan^2{\frac{k\pi}{2n}}$ 是以下多项式方程的根
\[\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(-x\right)^{n-r}=0.\]
根据 Vieta公式我们有
\[\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}.\]
进一步，我们有
\[\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}}= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.\]
证：
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} = \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} \\&= \sum\limits_{k = 1}^{2n - 1} {\frac{{{{\left( {{{\sin }^2}\frac{{k\pi }}{{2n}} + {{\cos }^2}\frac{{k\pi }}{{2n}}} \right)}^2}}}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} \\&= \sum\limits_{k = 1}^{2n - 1} {\left( {1 + 2{{\cot }^2}\frac{{k\pi }}{{2n}} + {{\cot }^4}\frac{{k\pi }}{{2n}}} \right)} \\&= 2n - 1 + 2\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}.\end{align*}
先考虑前者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {2n - k} \right)\pi }}{{2n}}} + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {n - k} \right)\pi }}{{2n}}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)}}{3}.\end{align*}
再考虑后者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {2n - k} \right)\pi }}{{2n}}} + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {n - k} \right)\pi }}{{2n}}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^4}\frac{{k\pi }}{{2n}}} = 2\left[ {{{\left( {\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} } \right)}^2} - 2\sum\limits_{1 \le i < j \le n - 1} {{{\tan }^2}\frac{{i\pi }}{{2n}}{{\tan }^2}\frac{{j\pi }}{{2n}}} } \right]\\&= 2\left[ {\frac{{{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - 2\frac{{\left( \begin{array}{c}2n\\5\end{array} \right)}}{{\left( \begin{array}{c}2n\\1\end{array} \right)}}} \right] \\&= \frac{{2{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - \frac{{2\left( {n - 2} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {2n - 1} \right)}}{{15}} \\&= \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {24{n^2} + 36n - 78} \right)}}{{135}} = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}}.\end{align*}
由此
\begin{align*}\Rightarrow \sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}} &= 2n - 1 + \frac{{4\left( {n - 1} \right)\left( {2n - 1} \right)}}{3} + \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}} \\&= 2n - 1 + \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {8{n^2} + 12n + 34} \right)}}{{45}} \\&= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.\end{align*}
特别地有
\[ \Rightarrow \sum\limits_{k = 1}^{999} {{{\csc }^4}\frac{{k\pi }}{{1000}}} = 1000009999989 = {10^{12}} + {10^7} - 11.\]
数字12：
\begin{align*}\int_0^\infty {\frac{{xdx}}{{\sinh \left( {\sqrt 3 x} \right)}}} &= \int_0^\infty {\frac{{xdx}}{{\frac{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}{2}}}} = \int_0^\infty {\frac{{2xdx}}{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}} \\&\underline{\underline {t = {e^{ - \sqrt 3 x}}}} \frac{2}{3}\int_0^1 {\frac{{\ln t}}{{{t^2} - 1}}dt} = \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt} - \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t + 1}}dt} \\&= \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt} - \frac{1}{3}\left[ {\ln t\ln \left( {t + 1} \right)} \right]_0^1 + \frac{1}{3}\int_0^1 {\frac{{\ln \left( {t + 1} \right)}}{t}dt} \\&= \frac{1}{3}\int_1^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt} - \frac{1}{3}\int_{ - 1}^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt} = \frac{1}{3}\mathrm{Li}_2\left( 1 \right) - \frac{1}{3}\mathrm{Li}_2\left( { - 1} \right) \\&= \frac{1}{3}\left( {\frac{{{\pi ^2}}}{6} + \frac{{{\pi ^2}}}{{12}}} \right) = \frac{{{\pi ^2}}}{{12}}\end{align*}
若干常见的傅里叶级数
关于Fourier级数，我们有以下几种常用表达式
\begin{align*}&x = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\cos \left[ {\left( {2n - 1} \right)x} \right]}}{{{{\left( {2n - 1} \right)}^2}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {nx} \right)}}{n}} \qquad 0 \le x < \pi \\&{x^2}=\frac{{{\pi ^2}}}{3}+4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{\cos \left( {nx} \right)}}{{{n^2}}}} \qquad - \pi \le x \le \pi \\&\cos \left( {ax} \right) = \frac{{\sin \left( {a\pi } \right)}}{\pi }\left( {\frac{1}{a} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2a}}{{{a^2} - {n^2}}}\cos \left( {nx} \right)} } \right)\qquad - \pi \le x \le \pi \\&\cot x = \frac{1}{x} + \sum\limits_{n = 1}^\infty {\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}}\qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\frac{1}{{\sin x}} = \frac{1}{x} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}} \qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\left| {\cos x} \right| = \frac{2}{\pi } + \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\left| {\sin x} \right| = \frac{2}{\pi } - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\cos \left( {nx} \right)}}{n}} = \ln \left( {2\cos \frac{x}{2}} \right)\qquad - \pi < x < \pi \\&\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {nx} \right)}}{n}} = - \ln \left( {2\sin \frac{x}{2}} \right)\qquad 0 < x < 2\pi \\&{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{k = 1}^\infty {\frac{{a\cos kx - k\sin kx}}{{{k^2} + {a^2}}}} } \right)\qquad x \in \left( {0,2\pi } \right),a \ne 0\end{align*}
根据这些高端级数我们有以下结论：
结论一：\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.\]
解：由Fourier级数
\[{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty {\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\quad \;\;\;{\mkern 1mu} x \in \left( {0,2\pi } \right),a \ne 0\]
将$\displaystyle f(x)=e^{ax}(0<x<2\pi)$延拓为$\displaystyle(-\infty,+\infty)$上以$\displaystyle 2\pi$为周期的函数$\displaystyle \widetilde{f}(x)$，则必须有
\[ \widetilde{f}(0)=\widetilde{f}(2\pi)=\frac{f(2\pi-0)+f(0+0)}{2}=\frac{e^{2\pi a}+e^{0\cdot a}}{2}=\frac{e^{2\pi a}+1}{2}.\]
由此得到
\[\widetilde{f}(x)= \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty{\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\qquad {\mkern 1mu} - \infty < x < + \infty \]
令$\displaystyle a=1,x=0$，我们立得
\[\frac{{{e^{2\pi }} + 1}}{2} = \frac{{{e^{2\pi }} - 1}}{\pi }\left( {\frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} } \right).\]
即
\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.\]
一般地，我们有
\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {a^2}}}} = \frac{\pi }{{2a}}\frac{{{e^{2\pi }} + 1}}{{{e^{2a\pi }} - 1}} - \frac{1}{{2{a^2}}}.\]
sinn/n和cosn/n类型的若干问题
题一：求极限
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} .\]
解：取$\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} $，则
\[f\left( {x + m} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + m} \right)} \right|} \left( {m \in {N^*}} \right).\]
\begin{align*}f\left( x \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^m {\left| {\cos \left( {k + x} \right)} \right|} + \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{s = 1}^n {\left| {\cos \left( {s + m + x} \right)} \right|} = f\left( {x + m} \right).\end{align*}
又
\[f\left( {x + 2\pi } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + 2\pi } \right)} \right|} = f\left( x \right).\]
所以
\begin{align}\label{MA1}f\left( x \right) = f\left( {x + m} \right) = f\left( {x + 2\pi } \right).\qquad(1)\end{align}
对于$\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} $，其中$\displaystyle \frac{1}{n}$单调且$\displaystyle\lim\limits_{n\rightarrow \infty}{\frac{1}{n}}=0,|\cos(k+x)|$在$x\in[0,1]$上一致有界，由狄利克雷判别法可知：函数$\displaystyle f(x)$一致收敛.同时（1）式说明周期可为任意整数，又可为$\displaystyle 2\pi$，那么必为常数函数，$\displaystyle f(x)\equiv C.$所以
\begin{align*}f\left( 0 \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = \int_0^1 {f\left( x \right)dx} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_0^1 {\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} dx} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_0^1 {\left| {\cos \left( {k + x} \right)} \right|dx} } \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_k^{k + 1} {\left| {\cos t} \right|dt} } \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_1^{n + 1} {\left| {\cos t} \right|dt} \\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{n}{\pi }} \right] \\&= \frac{2}{\pi }\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \frac{2}{\pi }\end{align*}
注：在区间$[1,n+1]$中，含有函数$\omega(t)=|\cos t|$的周期个数是$\left[ {\frac{n}{\pi }} \right]$，每个周期的积分值都是$2$，所以得到上述结果.
\[\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{n}{\pi } - \left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{{\left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}}} \right) = 1\]
其中$\left( {\frac{n}{\pi }} \right)\in(0,1).$
另解一：构建$|\cos k|,k\in\mathbb{N}^*$在$|\cos x|$上的随机分布，所以$|\cos k|$的均值就为
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = E\left( {\left| {\cos k} \right|} \right) = \frac{{\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} }}{\pi } = \frac{2}{\pi }.\]
另解二：令
\[f(x)=\int_0^x{|\cos t|dt}. \]
当$x\in[n\pi,(n+1)\pi)$时，
\[\int_0^{n\pi } {\left| {\cos t} \right|dt} \le f\left( x \right) < \int_0^{\left( {n + 1} \right)\pi } {\left| {\cos t} \right|dt} .\]
\[\Rightarrow 2n\leq f(x)<2(n+1).\]
由此得
\[\frac{{2n}}{{\left( {n + 1} \right)\pi }} < \frac{{f\left( x \right)}}{x} < \frac{{2\left( {n + 1} \right)}}{{n\pi }}.\]
记$S_n=\sum\limits_{k = 1}^n {\left| {\cos k} \right|}$，则有
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = \mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right)}}{x} = \frac{2}{\pi }.\]
题二：证明
\[\sum\limits_{n = 1}^\infty {\frac{{\sin n}}{n}} = \frac{{\pi - 1}}{2}.\]
证明：由Fourier级数
\[\frac{{\pi-x}}{2} = \sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} \qquad 0 < x < 2\pi\]
令$x=1$，我们立得
\[\sum\limits_{n = 1}^\infty {\frac{{\sin n}}{n}} = \frac{{\pi - 1}}{2}.\]
题三：证明
\[\sum\limits_{n = 1}^\infty {\frac{{\cos n}}{n}} = - \frac{1}{2}\ln (2 - 2\cos 1).\]
证明：由Fourier级数
\[\sum_{k=1}^\infty \frac{\cos(kx)}{k}=-\frac{1}{2}\ln(2-2\cos x)\qquad x\in\mathbb{R}. \]
令$x=1$，我们立得
\[\sum\limits_{n = 1}^\infty {\frac{{\cos n}}{n}} = - \frac{1}{2}\ln (2 - 2\cos 1).\]
傅里叶变换求积分函数
来自刚哥的虐心的积分题：
\begin{align*}&\int_0^\infty {\frac{{\cos tx}}{{1 + {t^2}}}} dt;\\&\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du} .\end{align*}
解：事实上，由Fourier变换公式
\begin{align*}{e^{ - x}} &= \frac{2}{\pi }\int_0^\infty {cos\left( {\lambda x} \right)d\lambda \int_0^\infty {{e^{ - u}}} cos\left( {\lambda u} \right)} du = \frac{2}{\pi }\int_0^\infty {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda } ;\\{e^{ - x}}\cos x &= \frac{2}{\pi }\int_0^\infty {\cos \left( {xu} \right)du} \int_0^\infty {{e^{ - t}}\cos t\cos \left( {ut} \right)dt} = \frac{2}{\pi }\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du}.\end{align*}
我们得到
\begin{align*}\int_0^\infty {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda } &= \frac{\pi }{2}{e^{ - x}};\\\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du} &= \frac{\pi }{2}{e^{ - x}}\cos x.\end{align*}
一般地，找到以下结论
\begin{align*}\int_0^\infty {\frac{{\cos xu}}{{{\beta ^2} + {u^2}}}du} &= \frac{\pi }{{2\beta }}{e^{ - \beta x}} \qquad x > 0,\beta > 0;\\\int_0^\infty {\frac{{u\sin xu}}{{{\beta ^2} + {u^2}}}du} &= \frac{\pi }{2}{e^{ - \beta x}}\qquad x > 0,\beta > 0;\\\int_0^\infty {\frac{{{x^{\mu - 1}}\sin \left( {ax} \right)}}{{{x^2} + 1}}dx} &= - {a^{2 - \mu }}\mathbf{\Gamma} \left( {\mu - 2} \right){}_1{F_2}\left( {1;\frac{{3 - \mu }}{2},\frac{{4 - \mu }}{2};\frac{{{a^2}}}{4}} \right)\mathrm{sign}\left( a \right)\sin \frac{{\mu \pi }}{2}\\&+ \frac{\pi }{2}\sec \frac{{\mu \pi }}{2}\sinh \left( a \right)\qquad{\mathop{\rm Im}\nolimits} a = 0, - 1 < \mathrm{Re}\mu < 3;\\\int_0^\infty {\frac{{{x^{\mu - 1}}\cos \left( {ax} \right)}}{{{x^2} + 1}}dx} &= \frac{\pi }{{2\sin \frac{{\mu \pi }}{2}}}\cosh a + \frac{1}{2}\cos \frac{{\mu \pi }}{2}\mathbf{\Gamma} \left( \mu \right)\\&\left[ {{e^{ - a + i\pi \left( {1 - \mu } \right)}}\gamma \left( {1 - \mu , - a} \right) - {e^a}\gamma \left( {1 - \mu ,a} \right)} \right]\qquad a > 0,0 < \mathrm{Re}\mu < 3;\\\int_0^\infty {\frac{{{x^{2\mu + 1}}\sin \left( {ax} \right)}}{{{x^2} + {b^2}}}dx} &= - \frac{\pi }{{2\cos \left( {\mu \pi } \right)}}{b^{2\mu }}\mathrm{sinh}\left( {ab} \right) + \frac{{\sin \left( {\mu \pi } \right)}}{{2{a^{2\mu }}}}\mathbf{\Gamma} \left( {2\mu } \right)\\&\left[ {{}_1{F_1}\left( {1;1 - 2\mu ;ab} \right) + {}_1{F_1}\left( {1;1 - 2\mu ; - ab} \right)} \right] \qquad a > 0, - \frac{3}{2} < \mathrm{Re}\mu < \frac{1}{2};\\\int_0^\infty {\frac{{{x^{2\mu + 1}}\cos \left( {ax} \right)}}{{{x^2} + {b^2}}}dx} &= - \frac{\pi }{{2\sin \left[ {\left( {\mu + \frac{1}{2}} \right)\pi } \right]}}{b^{2\mu + 1}}\cosh \left( {ab} \right) + \frac{{\cos \left[ {\left( {\mu + \frac{1}{2}} \right)\pi } \right]}}{{2{a^{2\mu + 1}}}}\mathbf{\Gamma} \left( {2\mu + 1} \right)\\&\left[ {{}_1{F_1}\left( {1; - 2\mu ;ab} \right) + {}_1{F_1}\left( {1; - 2\mu ; - ab} \right)} \right] \qquad a > 0, - 1 < \mathrm{Re}\mu < \frac{1}{2}.\end{align*}
