«(函数转换积分法) 求定积分 $\int _0^{+\infty }\frac{\ln \left(x\right)}{1+x^2}\mathrm{\:}dx$»
by pluvet on Dec 14, 2019

$$ \begin{align}&\int _0^{+\infty }\frac{\ln \left(x\right)}{1+x^2}\mathrm{\:}dx\\&=\int_{0}^{1} \frac{\ln x}{1+x^{2}} d x+\int_{1}^{+\infty} \frac{\ln x}{1+x^{2}} d x\\&=\int_{0}^{1}\left[\frac{\ln x}{1+x^{2}}+x^{-2} \frac{\ln \frac{1}{x}}{1+\frac{1}{x^{2}}}\right] d x\\&=\int_{0}^{1}\left[\frac{\ln x}{1+x^{2}}+\frac{\ln \frac{1}{x}}{1+x^{2}}\right] d x\\&=\int_{0}^{1} 0 d x\\&=0\end{align} $$

此处利用的原理:

$$ \int_0^\infty f(x)\mathrm dx=\int_0^1(f(x)+x^{-2}f(x^{-1}))\mathrm dx $$

函数的图像

2019-12-14T07:24:59.png

再来一个例子

$$ \begin{align} &\int^{+\infin}_{0}\frac{dx}{(x^2+1)(x^\pi+1)}\\ &=\int^{1}_{0}\frac{dx}{(x^2+1)(x^\pi+1)}\\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)\left(x^{2}+1\right)} d x+\int_{0}^{1} \frac{x^{-2}}{\left(x^{-2}+1\right)\left(x^{-2}+1\right)} d x\\ &=\int_{0}^{1} \frac{1}{x^{2}+1}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{-2}+1}\right) d x\\ \frac{1}{x^{2}+1}+\frac{1}{x^{-2}+1}=constant=1\\ &=\int_{0}^{1} \frac{1}{x^{2}+1} d x\\ &=\frac{\pi}{4} \end{align} $$

可以发现, 我们把 $\left[1, +\infin\right)$ 的空间压缩到了 $\left(0,1\right]$, 和芝诺悖论有异曲同工之妙.

添加新评论