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发表于 2021-1-11 22:44
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本帖最后由 永远 于 2022-11-28 18:11 编辑
正在一个一个验证,换个思路,我没必要纠结那个积分能不能求出来,我先绕过0这个发散点然后再回来反击它,不知道合不合法。这样避免逻辑尴尬。
\(\displaystyle\begin{gathered}
\int_0^1 {\frac{{\ln ( - \ln x)\ln (1 - x)}}{x}} dx = \int_0^1 {\ln ( - \ln x)\ln (1 - x)} d(\ln x) \hfill \\
\quad \,\;\;\,\;\,\underline{\underline {(令: - \ln x = t)}} \int_0^\infty {\ln t\ln (1 - {e^{ - t}})} dt \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = \int_0^\infty {\ln x\ln (1 - {e^{ - x}})} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = - \int_0^\infty {\ln x} \sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = - \int_0^\infty {\sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}\ln x} dx} \hfill \\
\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\,\, = - \mathop {\lim }\limits_{\lambda \to {0^ + }} \int_\lambda ^\infty {\sum\limits_{n = 1}^\infty {\frac{{{e^{ - nx}}}}{n}\ln x} dx} \hfill \\
\quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad\,\,\,\, \,\, = - \mathop {\lim }\limits_{\lambda \to {0^ + }} (\sum\limits_{n = 1}^\infty {\int_\lambda ^\infty {\frac{{{e^{ - nx}}}}{n}\ln xdx} } ) \hfill \\
\quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\, \; = - \sum\limits_{n = 1}^\infty {(\mathop {\lim }\limits_{\lambda \to {0^ + }} \int_\lambda ^\infty {\frac{{{e^{ - nx}}}}{n}\ln xdx} } ) \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad\; \; \;\;\; = - \sum\limits_{n = 1}^\infty {\int_0^\infty {\frac{{{e^{ - nx}}}}{n}\ln xdx} } \hfill \\
\quad \quad \quad \quad \quad \; = \sum\limits_{n = 1}^\infty {\frac{{\gamma + \ln n}}{{{n^2}}}} \hfill \\
\quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \;\; = \sum\limits_{n = 1}^\infty {\frac{\gamma }{{{n^2}}}} - {\left. {\frac{d}{{ds}}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} } \right|_{s = 2}} \hfill \\
\quad \quad \quad \quad \;\;\;\;\;= \frac{{\gamma {\pi ^2}}}{6} - \zeta '(2) \hfill \\
\end{gathered} \) |
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