※ 引述《JohnMash (Paul)》之銘言：
: ※ 引述《shooter31 (上揚的弧線)》之銘言：
: : For what values of the constants b and c will the following limit exist and
: : be equal to 1?
: : n x^3 + bx^2 + cx
: : lim ∫ ------------------- dx
: : n→∞ -n x^2 + x +1
: x^3 + bx^2 + cx = x(x^2+x+1)+d(x^2+x+1)+e(2x+1)+f
: then (x^3+bx^2+cx)/(x^2+x+1)=x+d+e(2x+1)/(x^2+x+1)+f/(x^2+x+1)
: ^^^^^^
: trick
: then d must be 0 (why?)
而且 f=-e 因為常數項為 0. 且∫(2x+1)/(x^2+x+1) dx = 0, 由-∞積至∞
: ∫[-n,n] 1/(x^2+x+1) dx
: = ∫[-n,n] 1/((x+1/2)^2+3/4) dx
: =∫[-n+1/2,n+1/2] 1/(u^2+3/4) du
: =2∫[0,n-1/2] 1/(u^2+3/4) du + ∫[n-1/2,n+1/2] 1/(u^2+3/4) du
d(arctan((2/√3)(x+1/2))/dx = (2/√3)/(1 + (4/3)(x+1/2)^2)
= (√3/2)/(x^2 + x +1)
故 ∫f/(x^2+x+1) dx = (2/√3)f arctan((2/√3)(x+1/2)) |x=-∞~∞
= (2π/√3)f = 1, f = (√3/2π) = -e
x^3 + bx^2 + cx = x(x^2+x+1) - (√3/2π)(2x+1) + (√3/2π)
= x^3 + x^2 + (1-√3/π)x, b = 1, c = (1 - √3/π).
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