一个有趣的反常积分

一个有趣的积分:$\displaystyle \int_{0}^{+\infty}\frac{(\ln x)^2}{1+x^2}\textrm{d}x=\frac{\pi^3}{8}.$
已邀请:
首先,我们把积分拆成两个部分:
$\displaystyle\int_0^{+\infty}\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx=\int_0^1\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx+\int_1^{+\infty}\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx.$

设$\displaystyle I_1=\int_0^1\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx$,$\displaystyle I_2=\int_1^{+\infty}\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx.$


先考虑$ I_2$,设$x=\frac{1}{t}$,则:
$\displaystyle I_2=\int_1^{0}\cfrac{\left[\textrm{ln}\left( \frac{1}{t} \right)\right]^2}{1+\left( \frac{1}{t}\right)^2}d\left( \frac{1}{t}\right)=
\int_0^1\cfrac{\left(\textrm{ln}t \right)^2}{1+t^2}dt=I_1.$

因此,只需计算$I_1$便可:

由于当$x\in\left(-1 ,1\right]$时,有$\displaystyle\cfrac{1}{1+x^2} =\sum_{k=0}^{+\infty\ }\left(-1 \right)^k x^{2k}$,
$\displaystyle I_1=\int_0^1\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx=\sum_{k=0}^{+\infty\ }\left(-1 \right)^k \int_0^1 x^{2k}\left(\textrm{ln}x \right)^2dx.$


考虑积分$\int_0^1 x^{2k}\left(\textrm{ln}x \right)^2dx$,运用分部积分法,
$\displaystyle\int_0^1 x^{2k}\left(\textrm{ln}x \right)^2dx=\cfrac{1}{2k+1}\int_0^1 \left(\textrm{ln}x \right)^2d(x^{2k+1})$
$\displaystyle=\cfrac{1}{2k+1}\left\{x^{2k+1}(\textrm{ln}x)^2 |_0^1-\int_0^1 x^{2k+1}d\left(\textrm{ln}x \right)^2\right\}$ $\cdot\cdot\cdot\cdot\cdot\cdot*$
$\displaystyle=\cfrac{1}{2k+1}\left\{-2\int_0^1 x^{2k}\textrm{ln}xdx \right\}.$

再运用一次分部积分法,得到:
$\displaystyle\int_0^1 x^{2k}\left(\textrm{ln}x \right)^2dx=\cfrac{(-2)(-1)}{{(2k+1)}^2}\int_0^1 x^{2k}dx=\cfrac{2}{{(2k+1)}^3}.$ $\cdot\cdot\cdot\cdot\cdot\cdot**$

带入原式:
$\displaystyle I_1=\sum_{k=0}^{+\infty\ }\left(-1 \right)^k \int_0^1 x^{2k}\left(\textrm{ln}x \right)^2dx$
$\displaystyle=2\sum_{k=0}^{+\infty\ }\cfrac{\left(-1 \right)^k}{{(2k+1)}^3}=2\left( \cfrac{\pi^3}{32}\right)=\cfrac{\pi^3}{16}.$ $\cdot\cdot\cdot\cdot\cdot\cdot***$

综上,
$\displaystyle\int_0^{+\infty}\cfrac{\left(\textrm{ln}x \right)^2}{1+x^2}dx=2I_1=2\left( \cfrac{\pi^3}{16}\right)=\cfrac{\pi^3}{8}.$

————————————————————————————
$*$:$\displaystyle x^{2k+1}(\textrm{ln}x)^2 |_0^1=0-\lim\limits_{x \to 0}x^{2k+1}(\textrm{ln}x)^2=-\lim\limits_{x \to +\infty}\cfrac{(-\textrm{ln}x)^2}{x^{2k+1}}=0.$


$**$:更一般的,若$m$是非负整数,则:
$\displaystyle\int_0^1\cfrac{\left(\textrm{ln}x \right)^m}{1+x^2}dx$
$\displaystyle=\sum_{k=0}^{+\infty\ }\left(-1 \right)^k\int_0^1x^{2k}\left(\textrm{ln}x \right)^mdx$
$\displaystyle=\left( -1\right)^m m!\sum_{k=0}^{+\infty\ }\cfrac{\left(-1 \right)^k}{{(2k+1)}^{m+1}}.$


$***$:考虑$x^3$在区间$(-\pi,\pi)$上的傅里叶级数:
$\displaystyle x^3 = \sum_{n=1}^{+\infty\ }a_n\textrm{sin}(nx)$,其中$\displaystyle a_n=\cfrac{2}{\pi}\int_0^\pi x^3\textrm{sin}(nx)dx.$

用分部积分法可得$\displaystyle a_n=\cfrac{2}{\pi}\left\{ \cfrac{\pi(6-\pi^2 n^2)\textrm{cos}(\pi n)}{n^3}+\cfrac{3(\pi^2n^2-2)\textrm{sin}(\pi n)}{n^4}\right\}.$

由于$n$是正整数,因此$\textrm{sin}(\pi n)=0$,$\textrm{cos}(\pi n)=(-1)^n$,所以:
$\displaystyle a_n=\cfrac{2}{\pi}\left\{ \cfrac{\pi(6-\pi^2 n^2)(-1)^n}{n^3}\right\}= \cfrac{2(6-\pi^2 n^2)(-1)^n}{n^3}.$

代入原式,得到:
$\displaystyle x^3 = \sum_{n=1}^{+\infty\ } \cfrac{2(6-\pi^2 n^2)(-1)^n}{n^3}\textrm{sin}(nx)$
$\displaystyle = 12\sum_{n=1}^{+\infty\ }\cfrac{(-1)^n\textrm{sin}(nx)}{n^3}-2\pi^2\sum_{n=1}^{+\infty\ }\cfrac{(-1)^n\textrm{sin}(nx)}{n}.$

令$x= \cfrac{\pi}{2}$. 当$n$是整数时,$\textrm{sin}\left(\frac{\pi (2n+1)}{2}\right)=(-1)^{n}$,$\textrm{sin}\left(\frac{\pi (2n)}{2}\right)=0$. 所以:

$\displaystyle \left( \cfrac{\pi}{2}\right)^3 = 12\sum_{n=0}^{+\infty\ }\cfrac{(-1)^{2n+1}\textrm{sin}\left(\frac{\pi (2n+1)}{2}\right)}{(2n+1)^3}-2\pi^2\sum_{n=0}^{+\infty\ }\cfrac{(-1)^{2n+1}\textrm{sin}\left(\frac{\pi (2n+1)}{2}\right)}{2n+1}$

$\displaystyle \cfrac{\pi^3}{8} = 12\sum_{n=0}^{+\infty\ }\cfrac{-(-1)^{n}}{(2n+1)^3}-2\pi^2\sum_{n=0}^{+\infty\ }\cfrac{-(-1)^{n}}{2n+1}$

$\displaystyle \cfrac{\pi^3}{8} = -12\sum_{n=0}^{+\infty\ }\cfrac{(-1)^{n}}{(2n+1)^3}+2\pi^2\left( \cfrac{\pi}{4}\right)$

$\displaystyle\therefore\sum_{k=0}^{+\infty\ }\cfrac{\left(-1 \right)^k}{{(2k+1)}^3}=\cfrac{\pi^3}{32}.$

要回复问题请先登录注册