求定积分cos nx/(cos x+cosh a)在[-pi,pi]的值

怎么证明$\displaystyle \frac1\pi\int_{-\pi}^{\pi}\frac{\cos nx}{\cos x+\cosh a}dx=\frac{4e^{a(1-n)}(-1)^n}{1-e^{2a}}$,其中$n\in\mathbb{N},a>0$
可否用含参变量来求,还是要用数学归纳法(只想作业本里少写几行)
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Eufisky - 熊熊

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PS:你这题目上少了个负号,囧!!!
注意到
\begin{align*}
&{a_{n + 1}} + {a_{n - 1}} = \frac{1}{\pi }\int_{ - \pi }^\pi {\frac{{\cos \left( {n + 1} \right)x + \cos \left( {n - 1} \right)x}}{{\cos x + \cosh a}}dx} \\
=&\frac{2}{\pi }\int_{ - \pi }^\pi {\frac{{\cos nx\cos x}}{{\cos x + \cosh a}}dx} = \frac{2}{\pi }\int_{ - \pi }^\pi {\cos nxdx} - \frac{{2\cosh a}}{\pi }\int_{ - \pi }^\pi {\frac{{\cos nx}}{{\cos x + \cosh a}}dx} \\
=& - 2\cosh a \cdot {a_n}.
\end{align*}
当$m>1$时,我们有
\begin{align*}
\int_0^\pi {\frac{{dx}}{{\cos x + m}}} & = \int_0^\pi {\frac{{dx}}{{\left( {m + 1} \right){{\cos }^2}\frac{x}{2} + \left( {m - 1} \right){{\sin }^2}\frac{x}{2}}}} \\
& = 2\int_0^{ + \infty } {\frac{{dt}}{{\left( {m + 1} \right) + \left( {m - 1} \right){t^2}}}} = \frac{\pi }{{\sqrt {{m^2} - 1} }}.
\end{align*}
因此
\begin{align*}
{a_0} &= \frac{1}{\pi }\int_{ - \pi }^\pi {\frac{1}{{\cos x + \cosh a}}dx} = \frac{2}{\pi }\int_0^\pi {\frac{1}{{\cos x + \cosh a}}dx} \\
&= \frac{2}{\pi } \cdot \frac{\pi }{{\sqrt {{{\cosh }^2}a - 1} }} = \frac{4}{{{e^a} - {e^{ - a}}}} = \frac{{ - 4{e^a}}}{{1 - {e^{2a}}}}\\
{a_1} &= \frac{1}{\pi }\int_{ - \pi }^\pi {\frac{{\cos x}}{{\cos x + \cosh a}}dx} = \frac{1}{\pi }\int_{ - \pi }^\pi {dx} - \frac{{\cosh a}}{\pi }\int_{ - \pi }^\pi {\frac{1}{{\cos x + \cosh a}}dx} \\
&= 2 - \cosh a \cdot {a_0} = \frac{4}{{1 - {e^{2a}}}}.
\end{align*}
由${a_{n + 1}} + {a_{n - 1}} = - 2\cosh a \cdot {a_n}$可知
\begin{align*}{a_{n + 1}} + {e^{ - a}}{a_n} = \left( { - {e^a}} \right)\left( {{a_n} + {e^{ - a}}{a_{n - 1}}} \right).\end{align*}

\begin{align*}{a_n} + {e^{ - a}}{a_{n - 1}} = {\left( { - {e^a}} \right)^{n - 1}}\left( {{a_1} + {e^{ - a}}{a_0}} \right) = 0 \Rightarrow \frac{{{a_n}}}{{{a_{n - 1}}}} = - {e^{ - a}}.\end{align*}
因此
\begin{align*}{a_n} = {\left( { - {e^{ - a}}} \right)^n}{a_0} = {\left( { - {e^{ - a}}} \right)^n} \cdot \frac{{ - 4{e^a}}}{{1 - {e^{2a}}}} = {\left( { - 1} \right)^{n + 1}}\frac{{4{e^{a\left( {1 - n} \right)}}}}{{1 - {e^{2a}}}}.\end{align*}

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