来自斯坦福轮回赛的有趣积分题目

另法： $I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos(x)+a^2\right)\ \text{d}x,;\ a>1$

设三角形ABC，边$a,b,c$所对的角为$\alpha,\beta,\gamma$

根据余弦定理有： $c^2=a^2+b^2-2ab\cos(\gamma)$

令$b=1,\gamma=x$

则：

$I(a) = \int_{0}^{\pi}\ln\left(c^2\right)\ \text{d}\gamma \ = 2\int_{0}^{\pi}\ln\left(c\right)\ \text{d}\gamma $

根据正弦定理有： $\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$

故：

$I(a) = 2\int_{0}^{\pi}\ln\left(a\ \frac{\sin(\gamma)}{\sin(\alpha)}\right)\ \text{d}\gamma \ = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\gamma $

又因为：$\gamma=\pi-\alpha-\beta$,$\text{d}\gamma=-\text{d}\alpha-\text{d}\beta$

当$\gamma\rightarrow 0:\alpha\rightarrow \pi,\beta\rightarrow 0$

当$\gamma\rightarrow \pi:\alpha\rightarrow 0,\beta\rightarrow 0$

故 $I(a) = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma\ -2\left[\int_{\pi}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\alpha)+\int_{0}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\beta)\right] \ = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\alpha = 2\pi \ln(a)$