求积分 \(\displaystyle\int_{0}^{1} \frac{\ln(-\ln x) \ln(1-x)}{x} dx\)
这个在怎么继续？？？\[\begin{gathered}
\int_0^1 {\frac{{\ln ( - \ln x)\ln (1 - x)}}{x}} dx = \int_0^1 {\ln ( - \ln x)\ln (1 - x)} d(\ln x) \hfill \\
\quad \quad \quad \quad \,\;\;\,\;\,\underline{\underline {( 令：- \ln x = t)}} \int_0^\infty{\ln t\ln (1 - {e^{ - t}})} dt \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, = \int_0^\infty{\ln x\ln (1 - {e^{ - x}})} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, =- \int_0^\infty{\ln x} \sum\limits_{n = 1}^\infty{\frac{{{e^{ - nx}}}}{n}} dx \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \, =- \int_0^\infty{\sum\limits_{n = 1}^\infty{\frac{{{e^{ - nx}}}}{n}\ln x} dx}\hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, =- \mathop {\lim }\limits_{\lambda\to {0^ + }} \int_\lambda ^\infty{\sum\limits_{n = 1}^\infty{\frac{{{e^{ - nx}}}}{n}\ln x} dx}\hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, =- \mathop {\lim }\limits_{\lambda\to {0^ + }} \sum\limits_{n = 1}^\infty{\int_\lambda ^\infty{\frac{{{e^{ - nx}}}}{n}\ln xdx} }\hfill \\
\end{gathered} \] 求极限\(\displaystyle\mathop {\lim }\limits_{\lambda\to {0^ + }} \sum\limits_{n = 1}^\infty{\int_\lambda ^\infty{\frac{{{e^{ - nx}}}}{n}\ln xdx} } \) 在附件里，能看清吗？可以用吗？ ShuXueZhenMiHu 发表于 2020-12-28 11:13
在附件里，能看清吗？可以用吗？
2楼那个极限怎么算，可有文献 \(\int_0^{\infty}\exp(-nx)\ln(x)dx\)对于\(\ln(x)\)做泰勒展开再积分即可
页:
[1]