请问这个含参量积分问题怎么求吖

\displaystyle\int_0^\pi \ln (1-2a\cos x+a^2)dx

最佳答案

\displaystyle I(a)=\int_0^\pi\ln(1-2a\cos x+a^2)dx\\{}\\ |a|<1,1-2a\cos x+a^2\geqslant1-2|a|+a^2=(1-|a|)^2>0\\{}\\ \therefore\ln(1-2a\cos x+a^2)\to continuity\\{}\\ \therefore\int_0^\pi\frac{-2\cos x+2a}{1-2a\cos x+a^2}dx=\frac{1}{a}\int_0^\pi\left(1+\frac{a^2-1}{1-2a\cos x+a^2}\right)dx\\{}\\ =\frac{\pi}{a}-\frac{1-a^2}{a(1+a^2)}\int_0^\pi\frac{dx}{1+\left(\frac{-2a}{1+a^2}\right)\cos x}=\frac{\pi}{a}-\frac{2}{a}\tan^{-1}\left(\frac{1+a}{1-a}\tan\frac{x}{2}\right)\bigg|_0^\pi=0\\{}\\ \left\{ \begin{aligned} I(a)=C_{onstant} && |a|<1\\ I(0)=0 && I(a)=0\\ \end{aligned} \right.\\{}\\ |a|>1,b=\frac{1}{a}\Rightarrow I(b)=0,|b|<1\\{}\\ I(a)=\int_0^\pi\ln\left(\frac{b^2-2b\cos x+1}{b^2}\right)dx=I(b)-2\pi\ln|b|=2\pi\ln|a|\\{}\\ |a|=1,I(1)=\int_0^\pi\ln2(1-\cos x)dx=\int_0^\pi\left(\ln4+2\ln\sin\frac{\pi}{2}\right)dx=0\\{}\\ \therefore I(-1)=0\\{}\\ I(a)=\left\{ \begin{aligned} 0, && |a|\leqslant1\\ 2\pi\ln|a|, && |a|>1 \end{aligned} \right.\\{}\\

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讨论数量: 1

\displaystyle I(a)=\int_0^\pi\ln(1-2a\cos x+a^2)dx\\{}\\ |a|<1,1-2a\cos x+a^2\geqslant1-2|a|+a^2=(1-|a|)^2>0\\{}\\ \therefore\ln(1-2a\cos x+a^2)\to continuity\\{}\\ \therefore\int_0^\pi\frac{-2\cos x+2a}{1-2a\cos x+a^2}dx=\frac{1}{a}\int_0^\pi\left(1+\frac{a^2-1}{1-2a\cos x+a^2}\right)dx\\{}\\ =\frac{\pi}{a}-\frac{1-a^2}{a(1+a^2)}\int_0^\pi\frac{dx}{1+\left(\frac{-2a}{1+a^2}\right)\cos x}=\frac{\pi}{a}-\frac{2}{a}\tan^{-1}\left(\frac{1+a}{1-a}\tan\frac{x}{2}\right)\bigg|_0^\pi=0\\{}\\ \left\{ \begin{aligned} I(a)=C_{onstant} && |a|<1\\ I(0)=0 && I(a)=0\\ \end{aligned} \right.\\{}\\ |a|>1,b=\frac{1}{a}\Rightarrow I(b)=0,|b|<1\\{}\\ I(a)=\int_0^\pi\ln\left(\frac{b^2-2b\cos x+1}{b^2}\right)dx=I(b)-2\pi\ln|b|=2\pi\ln|a|\\{}\\ |a|=1,I(1)=\int_0^\pi\ln2(1-\cos x)dx=\int_0^\pi\left(\ln4+2\ln\sin\frac{\pi}{2}\right)dx=0\\{}\\ \therefore I(-1)=0\\{}\\ I(a)=\left\{ \begin{aligned} 0, && |a|\leqslant1\\ 2\pi\ln|a|, && |a|>1 \end{aligned} \right.\\{}\\

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