第10章 · 第10-5-4题

[AI解答]

[AI解答]

**（1）** 设 $$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln\frac{1+a\cos x}{1-a\cos x}\cdot\frac{dx}{\cos x},\quad |a|<1. $$ 对参数 $a$ 求导（积分号下求导，满足一致收敛条件）： $$ I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{\partial}{\partial a}\left[\ln\frac{1+a\cos x}{1-a\cos x}\right]\frac{dx}{\cos x}. $$ 由于 $$ \frac{\partial}{\partial a}\ln\frac{1+a\cos x}{1-a\cos x} = \frac{\cos x}{1+a\cos x} - \frac{-\cos x}{1-a\cos x} = \frac{\cos x}{1+a\cos x} + \frac{\cos x}{1-a\cos x} = \frac{2\cos x}{1-a^{2}\cos^{2}x}. $$ 因此 $$ I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{2\cos x}{1-a^{2}\cos^{2}x}\cdot\frac{dx}{\cos x} =2\int_{0}^{\frac{\pi}{2}} \frac{dx}{1-a^{2}\cos^{2}x}. $$ 利用恒等式 $\cos^{2}x = \frac{1+\cos 2x}{2}$， $$ 1-a^{2}\cos^{2}x = 1-\frac{a^{2}}{2}(1+\cos 2x) = 1-\frac{a^{2}}{2} - \frac{a^{2}}{2}\cos 2x = \frac{2-a^{2}}{2} - \frac{a^{2}}{2}\cos 2x. $$ 于是 $$ I'(a)=2\int_{0}^{\frac{\pi}{2}} \frac{dx}{\frac{2-a^{2}}{2} - \frac{a^{2}}{2}\cos 2x} =4\int_{0}^{\frac{\pi}{2}} \frac{dx}{2-a^{2} - a^{2}\cos 2x}. $$ 令 $t=2x$，则 $dx = dt/2$，积分限 $0\to\pi$， $$ I'(a)=4\int_{0}^{\pi} \frac{dt/2}{2-a^{2} - a^{2}\cos t} =2\int_{0}^{\pi} \frac{dt}{2-a^{2} - a^{2}\cos t}. $$ 利用万能公式：令 $u=\tan\frac{t}{2}$，则 $\cos t = \frac{1-u^{2}}{1+u^{2}}$，$dt = \frac{2}{1+u^{2}}du$， 分母： $$ 2-a^{2} - a^{2}\frac{1-u^{2}}{1+u^{2}} = \frac{(2-a^{2})(1+u^{2}) - a^{2}(1-u^{2})}{1+u^{2}} = \frac{2-a^{2} + (2-a^{2})u^{2} - a^{2} + a^{2}u^{2}}{1+u^{2}} = \frac{2-2a^{2} + 2u^{2}}{1+u^{2}} = \frac{2(1-a^{2}+u^{2})}{1+u^{2}}. $$ 因此 $$ I'(a)=2\int_{0}^{\infty} \frac{\frac{2}{1+u^{2}} du}{\frac{2(1-a^{2}+u^{2})}{1+u^{2}}} =2\int_{0}^{\infty} \frac{du}{1-a^{2}+u^{2}}. $$ 这是一个标准积分： $$ \int_{0}^{\infty} \frac{du}{u^{2}+c^{2}} = \frac{\pi}{2c},\quad c>0. $$ 这里 $c=\sqrt{1-a^{2}}$，所以 $$ I'(a)=2\cdot \frac{\pi}{2\sqrt{1-a^{2}}} = \frac{\pi}{\sqrt{1-a^{2}}}. $$ 又因为 $I(0)=0$，积分得 $$ I(a)=\int_{0}^{a} \frac{\pi}{\sqrt{1-t^{2}}} dt = \pi \arcsin a. $$ 因此 $$ \boxed{I=\pi\arcsin a}. $$

**（2）** 设 $$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln\left(\cos^{2}x + a^{2}\sin^{2}x\right)dx,\quad a>0. $$ 对 $a$ 求导： $$ I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{2a\sin^{2}x}{\cos^{2}x + a^{2}\sin^{2}x} dx. $$ 分子分母同除以 $\cos^{2}x$，令 $t=\tan x$，则 $dx = \frac{dt}{1+t^{2}}$，积分限 $0\to\infty$， $$ I'(a)=\int_{0}^{\infty} \frac{2a t^{2}}{1+a^{2}t^{2}} \cdot \frac{dt}{1+t^{2}}. $$ 利用部分分式： $$ \frac{2a t^{2}}{(1+a^{2}t^{2})(1+t^{2})} = \frac{A}{1+a^{2}t^{2}} + \frac{B}{1+t^{2}}. $$ 通分比较分子： $$ 2a t^{2} = A(1+t^{2}) + B(1+a^{2}t^{2}). $$ 令 $t^{2}=-1$ 得 $ -2a = B(1-a^{2}) \Rightarrow B = \frac{-2a}{1-a^{2}}$。 令 $t^{2}=-\frac{1}{a^{2}}$ 得 $\frac{2a}{a^{2}} = A\left(1-\frac{1}{a^{2}}\right) \Rightarrow \frac{2}{a} = A\frac{a^{2}-1}{a^{2}} \Rightarrow A = \frac{2a}{a^{2}-1}$。 注意 $A = -B$，所以 $$ \frac{2a t^{2}}{(1+a^{2}t^{2})(1+t^{2})} = \frac{2a}{a^{2}-1}\left( \frac{1}{1+a^{2}t^{2}} - \frac{1}{1+t^{2}} \right). $$ 于是 $$ I'(a)=\frac{2a}{a^{2}-1}\int_{0}^{\infty} \left( \frac{1}{1+a^{2}t^{2}} - \frac{1}{1+t^{2}} \right) dt. $$ 分别积分： $$ \int_{0}^{\infty} \frac{dt}{1+a^{2}t^{2}} = \frac{1}{a}\cdot\frac{\pi}{2},\quad \int_{0}^{\infty} \frac{dt}{1+t^{2}} = \frac{\pi}{2}. $$ 所以 $$ I'(a)=\frac{2a}{a^{2}-1}\left( \frac{\pi}{2a} - \frac{\pi}{2} \right) = \frac{2a}{a^{2}-1}\cdot \frac{\pi}{2}\left( \frac{1}{a} - 1 \right) = \frac{\pi a}{a^{2}-1}\cdot \frac{1-a}{a} = \frac{\pi(1-a)}{a^{2}-1}. $$ 注意 $a^{2}-1 = (a-1)(a+1)$，而 $1-a = -(a-1)$，所以 $$ I'(a)=\frac{-\pi(a-1)}{(a-1)(a+1)} = -\frac{\pi}{a+1}. $$ 因此 $$ I(a) = -\pi \ln(a+1) + C. $$ 为确定常数 $C$，取 $a=1$： $$ I(1)=\int_{0}^{\frac{\pi}{2}} \ln(\cos^{2}x+\sin^{2}x)dx = \int_{0}^{\frac{\pi}{2}} \ln 1\,dx = 0. $$ 代入得 $$ 0 = -\pi \ln 2 + C \quad\Rightarrow\quad C = \pi\ln 2. $$ 因此 $$ I(a) = \pi\ln 2 - \pi\ln(a+1) = \pi\ln\frac{2}{a+1}. $$ 故 $$ \boxed{I=\pi\ln\frac{2}{a+1}}. $$

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