积分与级数一题

证明: $\int_0^{+\infty}\frac{1}{x}\int_0^x\frac{\cos(x-y)-\cos x}{y}dydx=\sum\limits_{n=1}^{+\infty}\frac{1}{n^2}$.
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Eufisky - 熊熊

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这题是MAA杂志里的征解题,以下解答是翻译而来的.
解.(翻译而来)令$f\left( {x,y} \right) = \frac{{\cos \left( {x - y} \right) - \cos x}}{y}$.对$x>0$,我们有\begin{align*}\int_0^x {f\left( {x,y} \right)dy} = \int_0^1 {\frac{{\cos \left( {1 - t} \right)x - \cos x}}{y}dt} = x\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\, dudt} .\end{align*}

因而对$R>0$,

\begin{align*}\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx} = \int_0^R {\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\,dudtdx} } .\end{align*}

而$|\sin ux|\leq1$,该三重积分是绝对收敛的.由Fubini定理可知积分能交换次序

\begin{align*}&\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx} = \int_0^1 {\int_{1 - t}^1 {\frac{1}{t}} \int_0^R {\sin ux\,dxdudt} } \\=& \int_0^1 {\frac{{1 - \cos Ru}}{u}\int_{1 - u}^1 {\frac{1}{t}} \,dtdu} \\= & - \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\,du} + \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du} .\end{align*}
我们知$|\ln (1-u)/u|\in L^1([0,1])$,由Riemann-Lebesgue引理可知
\begin{align*}\mathop {\lim }\limits_{R \to \infty } \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du} = 0.\end{align*}
由于${\sum\limits_{n = 1}^\infty {\frac{{{t^{n - 1}}}}{n}} }$一致收敛,故可逐项积分.因此我们有
\begin{align*}&\mathop {\lim }\limits_{R \to \infty } \int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)dydx} = - \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}du} \\= &\int_0^1 {\sum\limits_{n = 1}^\infty {\frac{{{t^{n - 1}}}}{n}} dt} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}.\end{align*}

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