$\int_0^\pi\ln(1-2\alpha\cos x+\alpha^2)dx$

hbghlyj

$f(\alpha)=\int_0^\pi\ln(1-2\alpha\cos x+\alpha^2)dx=2\pi\ln\abs\alpha$ for $\abs\alpha>1$
Mathworld
Woods, F. S. "Differentiation of a Definite Integral." §60 in Advanced Calculus: A Course Arranged with Special Reference to the Needs of Students of Applied Mathematics. Boston, MA: Ginn, pp. 141-144, 1926.

证明
$f(1)=\int_{0}^{\pi} \ln (2-2 \cos x) d x=0$
$\frac{\partial f}{\partial \alpha}=\int_{0}^{\pi} \frac{\partial}{\partial \alpha} \ln \left(1-2 \alpha \cos x+\alpha^{2}\right) d x=\int_{0}^{\pi} \frac{2(\alpha-\cos x)}{1-2 \alpha \cos x+\alpha^{2}} d x$
\begin{aligned}\int_{0}^{\pi} \frac{2(\alpha-\cos x)}{1-2 \alpha \cos x+\alpha^{2}} d x
&=\frac{1}{\alpha} \int_{0}^{\pi} \frac{2\left(\alpha^{2}-\alpha \cos x\right)}{1-2 \alpha \cos x+\alpha^{2}} d x
\\&=\frac{1}{\alpha} \int_{0}^{\pi}\left(\frac{2\left(\alpha^{2}-\alpha \cos x\right)}{1-2 \alpha \cos x+\alpha^{2}}+1-1\right) d x \\&=\frac{1}{\alpha} \int_{0}^{\pi}\left(\frac{2\left(\alpha^{2}-\alpha \cos x\right)}{1-2 \alpha \cos x+\alpha^{2}}-\frac{1-2 \alpha \cos x+\alpha^{2}}{1-2 \alpha \cos x+\alpha^{2}}+1\right) d x
\\&=\frac{1}{\alpha} \int_{0}^{\pi}\left(1-\frac{1-\alpha^{2}}{1+\alpha^{2}-2 \alpha \cos x}\right) d x
\\&=\frac{\pi}{\alpha}-\frac{1}{\alpha} \frac{1-\alpha^{2}}{1+\alpha^{2}} \int_{0}^{\pi}\left(\frac{1}{1-\frac{2 \alpha}{1+\alpha^{2}} \cos x}\right) d x\end{aligned}为去掉cos，我们可以进行变量替换$x=2 \arctan u,\cos x=\frac{1-u^{2}}{1+u^{2}},u=\tan \frac{x}{2}, d x=\frac{2 d u}{1+u^{2}}$.
\begin{aligned} \int_{0}^{\pi}\left(\frac{1}{1-\frac{2 \alpha}{1+\alpha^{2}} \cos x}\right) d x & =\int_{0}^{\infty} \frac{2 d u}{\left(1+u^{2}\right)\left(1-\frac{2 \alpha}{1+\alpha^{2}} \frac{1-u^{2}}{1+u^{2}}\right)} \\ & =\int_{0}^{\infty} \frac{2 d u}{\left(1-\frac{2 \alpha}{1+\alpha^{2}}\right)+\left(1+\frac{2 \alpha}{1+\alpha^{2}}\right) u^{2}} \\ & =\int_{0}^{\infty} \frac{2\left(1+\alpha^{2}\right) d u}{\left(1+\alpha^{2}-2 \alpha\right)+\left(1+\alpha^{2}+2 \alpha\right) u^{2}} \\ & =\int_{0}^{\infty} \frac{2\left(1+\alpha^{2}\right) d u}{(1-\alpha)^{2}+(1+\alpha)^{2} u^{2}} \\ & =\frac{2\left(1+\alpha^{2}\right)}{(1-\alpha)^{2}} \int_{0}^{\infty} \frac{d u}{1+\left(\frac{1+\alpha}{1-\alpha}\right)^{2} u^{2}}\end{aligned}我们快完成了。我们进行替换：$y=\frac{1+\alpha}{1-\alpha} u, d y=\frac{1+\alpha}{1-\alpha} d u$
注意，由于 $|\alpha|\ge1$ 和 $u\ge0$，y 的范围从零到负无穷大。所以：\begin{aligned}&\frac{2\left(1+\alpha^{2}\right)}{(1-\alpha)^{2}} \int_{0}^{\infty} \frac{d u}{1+\left(\frac{1+\alpha}{1-\alpha}\right)^{2} u^{2}} \\&=\frac{2\left(1+\alpha^{2}\right)}{(1-\alpha)(1+\alpha)} \int_{0}^{-\infty} \frac{d y}{1+y^{2}} \\&=\left.\frac{2\left(1+\alpha^{2}\right)}{1-\alpha^{2}} \arctan y\right|_{0} ^{-\infty} \\&=\frac{-\pi\left(1+\alpha^{2}\right)}{1-\alpha^{2}}\end{aligned}结合并利用$f(1) = 0$，我们可以完成积分的计算：\begin{aligned} \frac{\partial f}{\partial \alpha} & =\frac{2 \pi}{\alpha} \Longrightarrow f(\alpha)=2 \pi \ln |\alpha|+C_{0} \\ f(1) & =2 \pi \ln (1)+C_{0}=0 \Longrightarrow C_{0}=0 \\ \therefore f(\alpha) & =\int_{0}^{\pi} \ln \left(1-2 \alpha \cos x+\alpha^{2}\right) d x=2 \pi \ln |\alpha|\end{aligned}

Czhang271828

注意到对 $z\in[0,\pi]$, 有全纯函数
\[\ln (\alpha-e^{iz})+\ln (\alpha-e^{-iz})=\ln (\alpha^2-2\alpha\cos z+1).\]
用一下全纯函数的平均值原理显然.

hbghlyj

Czhang271828 发表于 2023-4-11 15:41
用一下全纯函数的平均值原理显然

Apply Gauss's Mean-Value Theorem to $\ln (\alpha-e^{iz})$ at $z=0$
$$\frac1{2\pi}\int_0^{2\pi}\ln (\alpha-e^{iz})dz=\ln\alpha$$