问题及解答

P. 240,  习题 5.4

1.  讨论下列反常积分的敛散性, 如果收敛, 求反常积分的值:

(2)    $\displaystyle\int_{e}^{+\infty}\dfrac{\mathrm{d}x}{x\ln^2 x}$

2.  讨论下列无界函数的反常积分的敛散性, 如果收敛, 求反常积分的值:

(7)    $\displaystyle\int_{0}^{1}\dfrac{\ln x}{(1-x)^2}\mathrm{d}x$

1.  (2)

\[
\int_{e}^{a}\frac{\mathrm{d}x}{x\ln^2 x}=\int_{e}^{a}\frac{1}{\ln^2 x}\mathrm{d}\ln x=(-\frac{1}{\ln x})\biggr|_{x=e}^{a}=1-\frac{1}{\ln a}
\]

因此,

\[
\int_{e}^{+\infty}\frac{\mathrm{d}x}{x\ln^2 x}=\lim_{a\rightarrow+\infty}\int_{e}^{a}\frac{\mathrm{d}x}{x\ln^2 x}=\lim_{a\rightarrow+\infty}(1-\frac{1}{\ln a})=1
\]

即反常积分收敛, 且等于 1.

2.  (7)

\[
\begin{split}
\int_{\varepsilon}^{1-\varepsilon}\frac{\ln x}{(1-x)^2}\mathrm{d}x&=\int_{\varepsilon}^{1-\varepsilon}\ln x\mathrm{d}\frac{1}{1-x}\\
&=\frac{\ln x}{1-x}\biggr|_{\varepsilon}^{1-\varepsilon}-\int_{\varepsilon}^{1-\varepsilon}\frac{1}{1-x}\mathrm{d}\ln x\\
&=\Bigl(\frac{\ln(1-\varepsilon)}{1-(1-\varepsilon)}-\frac{\ln\varepsilon}{1-\varepsilon}\Bigr)+\int_{\varepsilon}^{1-\varepsilon}\frac{1}{x-1}\cdot\frac{1}{x}\mathrm{d}x\\
&=\frac{\ln(1-\varepsilon)}{\varepsilon}-\frac{\ln\varepsilon}{1-\varepsilon}+\int_{\varepsilon}^{1-\varepsilon}(\frac{1}{x-1}-\frac{1}{x})\mathrm{d}x\\
&=\frac{\ln(1-\varepsilon)}{\varepsilon}-\frac{\ln\varepsilon}{1-\varepsilon}+\Bigl(\ln(1-x)-\ln x\Bigr)\biggr|_{\varepsilon}^{1-\varepsilon}\\
&=\frac{\ln(1-\varepsilon)}{\varepsilon}-\frac{\ln\varepsilon}{1-\varepsilon}+\bigl[\ln(1-(1-\varepsilon))-\ln(1-\varepsilon)\bigr]-\bigl[\ln(1-\varepsilon)-\ln\varepsilon\bigr]\\
&=\frac{\ln(1-\varepsilon)}{\varepsilon}-\frac{\ln\varepsilon}{1-\varepsilon}+2\ln\varepsilon-2\ln(1-\varepsilon)\\
&=(\frac{1}{\varepsilon}-2)\ln(1-\varepsilon)+(2-\frac{1}{1-\varepsilon})\ln\varepsilon
\end{split}
\]

注意到

\[
(\frac{1}{\varepsilon}-2)\ln(1-\varepsilon)=\frac{\ln(1-\varepsilon)}{\varepsilon}-2\ln(1-\varepsilon)\rightarrow -1\quad(\text{当}\ \varepsilon\rightarrow 0^+)
\]

而

\[
(2-\frac{1}{1-\varepsilon})\ln\varepsilon\rightarrow -\infty\quad(\text{当}\ \varepsilon\rightarrow 0^+)
\]

故原反常积分发散.