这道题来自MAA的杂志The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), pp. 500-507,可以参考链接http://www.jstor.org/stable/10.4169/amer.math.monthly.122.5.500?seq=1#page_scan_tab_contents
求\[\int_0^\infty {\frac{1}{x}dx} \int_0^x {\frac{{\cos \left( {x - y} \right) - \cos x}}{y}dy} .\]
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\[\int_0^x {f\left( {x,y} \right)dy} = \int_0^1 {\frac{{\cos \left( {1 - t} \right)x - \cos x}}{y}dt} = x\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\, dudt} .\]
因而对$R>0$,
\[\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx} = \int_0^R {\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\,dudtdx} } .\]
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
\[\mathop {\lim }\limits_{R \to \infty } \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du} = 0.\]
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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