# 积分与级数一题

Eufisky - 熊熊

\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*}

\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*}

\begin{align*}\mathop {\lim }\limits_{R \to \infty } \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du} = 0.\end{align*}

\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*}