变量替换巧获积分关系
题源:自动内卷人(来自Q群-卷明白)
证明: $\int_{0}^{1} \frac{\ln (1+x)}{x} d x=\frac{\pi^{2}}{12}$, 并由此计算 $\int_{0}^{+\infty} \arctan x \cdot \ln \left(1+\frac{1}{x^{2}}\right) d x$.
前半部分:略
后半部分:
计算$\int_{0}^{+\infty} \arctan x \cdot \ln \left(1+\frac{1}{x^{2}}\right) d x$并不算很困难,但由前半部分证明后半部分存在一定难度。
相似的题:
SEEMOUS 2016 problem 4:
整数$n \ge 1$。设$$I_n=\int_0^{\infty}\frac{\arctan x}{(1+x^2)^n} dx.$$
证明
a) $\sum_{n=1}^{\infty}\frac{I_n}{n}=\frac{\pi ^2}{6}$
b)$\int_0^{\infty}\arctan\cdot \ln(1+\frac{1}{x^2})dx=\frac{\pi ^2}{6}$
$\int_0^\infty \arctan(x) \ln \left(1 + \frac{1}{x^2}\right) dx$
$=\int_0^\infty \ln \left(1 + \frac{1}{x^2}\right) d(xarctanx-\frac{1}{2}ln(1+x^2))$
$= \int_0^\infty \left[\frac{2 \arctan{x}}{1+x^2} - \frac{\ln(1+x^2)}{x (1+x^2)}\right] dx $
$\hspace{-11pt}\stackrel{x = \sqrt{\mathrm{e}^t - 1}}{=} \frac{\pi^2}{4} - \frac{1}{2} \int_0^\infty \frac{t}{\mathrm{e}^t - 1} \mathrm{d} t $
设:
$I=\int_{0}^{\infty} \frac{x}{e^{x}-1} d x$
则:
$I=\int_{0}^{\infty} \frac{x}{e^{x}-1} d x=-\int_{-\infty}^{0} \frac{-y}{e^{-y}-1} d y=\int_{-\infty}^{0} \frac{y \cdot d^{y}}{y-1} d y=\int_{0}^{1} \frac{u \cdot \ln (u)}{u \cdot(u-1)} d u$
$I=\int_{0}^{1} \frac{\ln (u)}{u-1} d u=4 \int_{0}^{1} \frac{v \cdot \ln (v)}{v^{2}-1} d \nu=\int_{0}^{1} \frac{u \cdot \ln (u)}{u^{2}-1} d u+\int_{0}^{1} \frac{\ln (u)}{u^{2}-1} d u$ (1)
由此说明:
$\int_{0}^{1} \frac{\ln (x)}{x^{2}-1} d x=3\int_{0}^{1} \frac{x\ln (x)}{x^{2}-1} d x$ (2)
除此之外,还有一个有趣的等式:
$ \int_{0}^{1} \frac{\ln (u)}{u^{2}-1} d u$
$=- \cdot \int_{\infty}^{1} \frac{\left(\frac{1}{w}\right)^{2} \cdot \ln \left(\frac{1}{w}\right)}{\left(\frac{1}{w}\right)^{2}-1} d w$
$= \int_{1}^{\infty} \frac{\ln (w)}{w^{2}-1} d v$
$= \int_{1}^{\infty} \frac{\ln (u)}{u^{2}-1} d u$
设$J=\int_{0}^{1} \frac{\ln (x+1)}{x} d x$,由题意我们知道$J=\frac{\pi^{2}}{12}$.
$J=\int_{0}^{1} \frac{\ln (x+1)}{x} d x$
$\hspace{-11pt}\stackrel{分部积分}{=} -\int_{0}^{1} \frac{\ln (x)}{x+1} d x$
$=-\int_{0}^{1} \frac{(x-1) \cdot \ln (x)}{(x-1) \cdot(x+1)} d x$
$=\int_{0}^{1} \frac{\ln (x)}{x^{2}-1} d x-\int_{0}^{1} \frac{x \cdot \ln (x)}{x^{2}-1} d x$
代入(2)得:
$\quad \int_{0}^{1} \frac{\ln (x)}{x^{2}-1} d x=\frac{\pi^{2}}{8} \quad \int_{0}^{1} \frac{x \ln (x)}{x^{2}-1} d x=\frac{\pi^{2}}{24}$
再代入(1)得:$I=\frac{\pi^{2}}{6}$
故原式$=\frac{\pi^{2}}{4}-\frac{1}{2}*I=\frac{\pi^{2}}{6}$