无穷区间上：求和号、极限号，积分号交换经典案例及苦逼疑问

永远

开篇以一个不起眼的例题开始：

计算：\(\displaystyle\int_0^\infty  {\frac{x}{{{e^x} - 1}}} dx\)

                                                                              

永远

\[\begin{gathered}
  \int_0^{ + \infty } {\frac{x}{{{e^x} - 1}}} dx = \int_0^{ + \infty } {\frac{{x{e^{ - x}}}}{{1 - {e^{ - x}}}}} dx \\
   = \int_0^{ + \infty } {\frac{{xd{e^{ - x}}}}{{{e^{ - x}} - 1}}}  \\
   = \int_0^{ + \infty } {xd\ln (1 - {e^{ - x}})}  \\
   = \left. {x\ln (1 - {e^{ - x}})} \right|_0^{ + \infty } - \int_0^{ + \infty } {\ln (1 - {e^{ - x}})d} x \\
   =  - \int_0^{ + \infty } {\ln (1 - {e^{ - x}})d} x \\
   =  - \int_0^{ + \infty } {\sum\limits_{n = 1}^\infty  {\frac{{{{( - 1)}^{n - 1}}}}{n}{{( - {e^{ - x}})}^n}} d} x \\
   = \int_0^{ + \infty } {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}}}{n}} d} x \\
   = \mathop {\lim }\limits_{\lambda  \to  + 0} \int_\lambda ^{ + \infty } {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - n\;x}}}}{n}} } dx \\
   = \mathop {\lim }\limits_{\lambda  \to  + 0} \sum\limits_{n = 1}^\infty  {\int_\lambda ^{ + \infty } {\frac{{{e^{ - n\;x}}}}{n}} dx}  \\
   = \mathop {\lim }\limits_{\lambda  \to  + 0} \sum\limits_{n = 1}^\infty  {\left. {( - \frac{{{e^{ - n\;x}}}}{{{n^2}}})} \right|_\lambda ^{ + \infty }}  \\
   = \mathop {\lim }\limits_{\lambda  \to  + 0} \sum\limits_{n = 1}^\infty  {\frac{{{e^{ - n\;\lambda }}}}{{{n^2}}}}  \\
   = \sum\limits_{n = 1}^\infty  {\mathop {\lim }\limits_{\lambda  \to  + 0} \frac{{{e^{ - n\;\lambda }}}}{{{n^2}}}}  \\
   = \sum\limits_{n = 1}^\infty  {\frac{{{e^{ - n\;0}}}}{{{n^2}}}}  \\
   = \sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}  \\
   = \frac{{{\pi ^2}}}{6} \\
\end{gathered} \]

永远

经典案例重要部分：

\(\displaystyle\boxed{\int_0^{ + \infty } {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - n\;x}}}}{n}} } dx = \mathop {\lim }\limits_{\lambda  \to {0^ + }} \int_\lambda ^{ + \infty } {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - n\;x}}}}{n}} } dx = \mathop {\lim }\limits_{\lambda  \to {0^ + }} \sum\limits_{n = 1}^\infty  {\int_\lambda ^{ + \infty } {\frac{{{e^{ - n\;x}}}}{n}} dx} }\)

永远

基础理论依据（转载之陆老师的贴子）：

永远

下面看另外一个类似的例子：

计算定积分 \(\displaystyle\int_{0}^{1} \frac{\ln(-\ln x) \ln(1-x)}{x} dx\)

永远

这是我给的过程，我写完后认为我后面是错的：

\(\displaystyle\begin{gathered}
  \int_0^1 {\frac{{\ln ( - \ln x)\ln (1 - x)}}{x}} dx = \int_0^1 {\ln ( - \ln x)\ln (1 - x)} d(\ln x) \hfill \\
\quad \,\;\;\,\;\,\underline{\underline {(令： - \ln x = t)}} \int_0^\infty  {\ln t\ln (1 - {e^{ - t}})} dt \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad \, = \int_0^\infty  {\ln x\ln (1 - {e^{ - x}})} dx \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad \, =  - \int_0^\infty  {\ln x} \sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}}}{n}} dx \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad \, =  - \int_0^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}}}{n}\ln x} dx}  \hfill \\
  \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\,\, =  - \mathop {\lim }\limits_{\lambda  \to {0^ + }} \int_\lambda ^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}}}{n}\ln x} dx}  \hfill \\
  \quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad\,\,\,\, \,\, =  - \mathop {\lim }\limits_{\lambda  \to {0^ + }} (\sum\limits_{n = 1}^\infty  {\int_\lambda ^\infty  {\frac{{{e^{ - nx}}}}{n}\ln xdx} } ) \hfill \\
\quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\, \; =  - \sum\limits_{n = 1}^\infty  {(\mathop {\lim }\limits_{\lambda  \to {0^ + }} \int_\lambda ^\infty  {\frac{{{e^{ - nx}}}}{n}\ln xdx} } ) \hfill \\
\quad \quad \quad \quad \quad \quad \quad \quad\; \; \;\;\; =  - \sum\limits_{n = 1}^\infty  {\int_0^\infty  {\frac{{{e^{ - nx}}}}{n}\ln xdx} }  \hfill \\
     \quad \quad \quad \quad \quad \; = \sum\limits_{n = 1}^\infty  {\frac{{\gamma  + \ln n}}{{{n^2}}}}  \hfill \\
    \quad\quad\quad \quad \quad \quad \quad \quad \quad \quad \;\; = \sum\limits_{n = 1}^\infty  {\frac{\gamma }{{{n^2}}}}  - {\left. {\frac{d}{{ds}}\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} } \right|_{s = 2}} \hfill \\
   \quad \quad \quad \quad \;\;\;\;\;= \frac{{\gamma {\pi ^2}}}{6} - \zeta '(2) \hfill \\
\end{gathered} \)

永远

楼主认为：\(\displaystyle\int_0^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}} } dx = \mathop {\lim }\limits_{\lambda  \to {0^ + }} \sum\limits_{n = 1}^\infty  {\int_\lambda ^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}dx} } \)


e老师认为：\(\displaystyle\int_0^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}} } dx = \sum\limits_{n = 1}^\infty  {\int_0^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}dx} } \)

永远

可e老师坚持认为：\(\displaystyle\boxed{\int_0^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}} } dx = \sum\limits_{n = 1}^\infty  {\int_0^\infty  {\frac{{{e^{ - nx}}\ln x}}{n}dx} } } \)

下面是他的分析：

题：试证\(\;\;\displaystyle\int_0^{\infty}\big(\sum_{n=1}^{\infty}{\small\frac{\ln x}{n}}e^{-nx}\big)dx =\sum_{n=1}^{\infty}\int_0^{\infty}{\small\frac{\ln x}{n}}e^{-nx}dx\).
证：设\(\;0< \alpha< \lambda< \infty,\;\)据 Cauchy-Schwarz不等式,\(\qquad\displaystyle\bigg(\int_{\alpha}^{\lambda}e^{-nx}|\ln x| dx\bigg)^2\le\int_{\alpha}^{\lambda}e^{-2nx}dx\int_{\alpha}^{\lambda}\ln^2 x dx\)\(\qquad\displaystyle=\small\frac{e^{-2n\alpha}-e^{-2n\lambda}}{2n}(\varphi^2(\lambda)-\varphi^2(\alpha))\quad(\varphi^2(x)=x(1+\ln^2(x/e)))\)
\((1)\quad\displaystyle 0<\overset{\,}{\small\sum_{n=1}^{\infty}\int_0^{\lambda}\frac{1}{n}}e^{-nx}|\ln x|dx <\varphi(\lambda)\zeta({\scriptsize\frac{3}{2}});\)\(\because\quad e^{-x/2}\ln x< e\;\;(x\ge e),\displaystyle\int_{\eta}^{\infty}e^{-nx}\ln x dx< e\int_{\eta}^{\infty}e^{-(n-1/2)x}dx\)
\((2)\quad\displaystyle{\small\sum_{n=1}^{\infty}\int_{\eta}^{\infty}\frac{1}{n}}e^{-nx}\ln xdx<{\small\sum_{n=1}^{\infty}\frac{e^{-(n+1/2)\eta}}{n(n-1/2)}}< 2e^{-\eta}\zeta(2);\quad\)已知
\((3)\quad\displaystyle{f(x)=\small\sum_{n=1}^{\infty}\frac{\ln x}{n}}e^{-nx}dx\;\)在\(x\ge a>0\,\)一致收敛且\({\displaystyle\small\int_0^{\infty}}f\,\)收敛.\(\therefore\quad\displaystyle\left|\sum_{n=1}^{N}\int_0^{\infty}\frac{\ln x}{n}e^{-nx}dx-\int_0^{\infty}f(x)dx\right|\)
\(\qquad\qquad\le\displaystyle\left|\sum_{n=1}^{N}\int_0^{\lambda}\frac{\ln x}{n}e^{-nx}dx-\int_0^{\lambda}f(x)dx\right|\)
\(\qquad\qquad\quad+\displaystyle\left|\int_{\lambda}^{\eta}\big(\sum_{n=1}^{N}\frac{\ln x}{n}e^{-nx}-f(x)\big)dx\right|\)
\(\qquad\qquad\quad+\displaystyle\left|\sum_{n=1}^{N}\int_{\eta}^{\infty}\frac{\ln x}{n}e^{-nx}dx-\int_{\eta}^{\infty}f(x)dx\right|\)
任给\(\,\varepsilon>0,\)存在\(\,\lambda>0,\,\eta>e\,\)使

\(\qquad\displaystyle{\small\int_0^{\lambda}|f|,\;\int_{\eta}^{\infty}f},\;\varphi(\lambda)\zeta({\scriptsize\frac{3}{2}}),\,2e^{-\eta}\zeta(2)\in(0,\varepsilon/6)\)
存在\(\;N\in\mathbb{N}\,\)使
\(\;\displaystyle\left|\sum_{n=1}^m\frac{\ln x}{n}e^{-nx}-f(x)\right|<\frac{\varepsilon}{3(\lambda+\eta)}\;\;(\forall x\in[\lambda,\eta],\forall m>N)\)\(\therefore\quad\displaystyle\int_0^{\infty}\big(\sum_{n=1}^{\infty}{\small\frac{\ln x}{n}}e^{-nx}\big)dx =\sum_{n=1}^{\infty}\int_0^{\infty}{\small\frac{\ln x}{n}}e^{-nx}dx\quad\small\square\)

永远

下面是陆老师对楼上案例部分细节分析

elim

论证让楼主崩溃，那就玩“认为”？