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发表于 2020-12-24 00:39
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本帖最后由 永远 于 2022-11-28 16:53 编辑
陆老师好,这个在怎么继续
\[\begin{gathered}
\int_0^1 {\frac{{\ln ( - \ln x)\ln (1 - x)}}{x}} dx = \int_0^1 {\ln ( - \ln x)\ln (1 - x)} d(\ln x) \hfill \\
\quad \quad \quad \quad \,\;\;\,\;\,\underline{\underline {( 令:- \ln x = t)}} \int_0^\infty {\ln t\ln (1 - {e^{ - t}})} dt \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = \int_0^\infty {\ln x\ln (1 - {e^{ - x}})} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = - \int_0^\infty {\ln x} \sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = - \int_0^\infty {\sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}\ln x} dx} \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, = - \mathop {\lim }\limits_{\lambda \to {0^ + }} \int_\lambda ^\infty {\sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}\ln x} dx} \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, = - \mathop {\lim }\limits_{\lambda \to {0^ + }} \sum\limits_{n = 1}^\infty {\int_\lambda ^\infty {\frac{{{e^{ - nx}}}}{n}\ln xdx} } \hfill \\
\end{gathered} \] |
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