一个很好的积分题

Eufisky posted @ 2015年8月22日 00:54 in 数学分析 with tags 积分计算 , 1084 阅读

背景是这个：

然后郝XX跟我说了下他的求法，还是挺有意思的！

求

\[\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} .\]

由Euler公式,我们有

\begin{align*}&{e^{y\cos x}}\cos \left( {y\sin x} \right) + i{e^{y\cos x}}\sin \left( {y\sin x} \right) = {e^{y\cos x}} \cdot {e^{iy\sin x}}\\=& {e^{y\cos x + iy\sin x}} = {e^{y\left( {\cos x + i\sin x} \right)}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} \\=& \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}\left( {\cos nx + i\sin nx} \right)} \\= &\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} + i\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\end{align*}

因此有

\[{e^{y\cos x}}\cos \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} ,{e^{y\cos x}}\sin \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\]

在第一个式子中令$y=1$,我们有

\[{e^{\cos x}}\cos \left( {\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} .\]

因此我们有

\begin{align*}&\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} = \int_0^{2\pi } {\left( {\cos nx \cdot \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} } \right)dx} \\=& \int_0^{2\pi } {\left( {\cos nx \cdot \frac{{\cos nx}}{{n!}}} \right)dx} = \left\{ \begin{array}{l}\frac{\pi }{{n!}},n \ge 1\\2\pi ,n = 0\end{array} \right..\end{align*}

而对于无穷乘积\[\prod\limits_{n = 3}^\infty {\cos \frac{\pi }{{n!}}} \approx 0.858314.\]也就是管理员一分钟都不能禁他！！！

事实上,我们有

\begin{align*}&\int_0^\pi {{e^{p\cos x}}\cos \left( {p\sin x} \right)\cos qxdx} = \int_0^\pi {{e^{p\cos x}}\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{ip\sin x}}} \right)dx} \\= &\int_0^\pi {\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{ip\sin x + p\cos x}}} \right)dx} = \int_0^\pi {\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{p{e^{ix}}}}} \right)dx} \\= &\int_0^\pi {\cos qx{\mathop{\rm Re}\nolimits} \left( {\sum\limits_{k = 0}^\infty {\frac{{{p^k}{e^{ikx}}}}{{k!}}} } \right)dx} = \sum\limits_{k = 0}^\infty {\frac{{{p^k}}}{{k!}}\int_0^\pi {\cos qx\cos kxdx} } \\= &\frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{{{p^k}}}{{k!}}\left[ {\frac{{\sin \left( {k - q} \right)x}}{{k - q}} - \frac{{\sin \left( {k + q} \right)x}}{{k + q}}} \right]} _0^\pi = \frac{\pi }{2}\frac{{{p^q}}}{{q!}}.\end{align*}