题目的积分上限是π下限是0设f(a)=∫上限π下限是0 ln（1-2acosx+a^2）dx，证明（1）f（-

1-2acosx+a^2≥0恒成立
(1)令t=π-x
∫[0,π]ln(1-2acosx+a^2)dx=∫[0,π]ln(1+2acosx+a^2)dt
所以F(-a)=F(a)
(2)不好证,但可以通过求导转化
2F'(a)=2aF'(a^2)
被积函数对a求偏导
F'(a)=∫[0,π](2a-2cosx)/(1-2acosx+a^2)dx
=∫[0,π][1/a+(a^2-1)/a*1/(1-2acosx+a^2)]dx
万能公式代换tan(x/2)=t
=π/a+(a^2-1)/a*∫[0,+∞]2/(1+t^2)*1/[1-2a(1-t^2)/(1+t^2)+a^2]dt
=π/a+(a^2-1)/a*∫[0,+∞]2/[(a^2+1)t^2+(a-1)^2]dt
=π/a+1/a*(a^2-1)/|a^2-1|*2arctan(x*|a+1|/|a-1|)[0,+∞]
=π/a+1/a*(a^2-1)/|a^2-1|*π
1.a^2>1,原式2π/(a)
2.a=±1,原式±2π
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