一道定积分的题,已知∫[0,1] [(e^x)/(1+x)]dx=A,求∫[a-1,a] [(e^(-x)/(x-a-1
2个回答

∫(0→1) e^x/(1 + x) dx = A

= ∫ e^(x + 1 - 1)/(1 + x) dx

= ∫ e^(x + 1) · e^(- 1)/(x + 1) d(x + 1)

= (1/e)∫ e^(x + 1)/(x + 1) d(x + 1)

令u = x + 1,du = d(x + 1)

当x = 0,u = 1,当x = 1,u = 2

= (1/e)∫(1→2) e^u/u du

= (1/e)∫(1→2) e^x/x dx = A

∫(a - 1→a) e^(- x)/(x - a - 1) dx

= ∫ e^(a - x + 1 - a - 1)/[- (a - x + 1)] dx

= ∫ e^(a - x + 1) · e^(- a - 1)/(a - x + 1) d(a - x + 1)

= e^(- a - 1) · ∫ e^(a - x + 1)/(a - x + 1) d(a - x + 1)

令v = a - x + 1,dv = d(a - x + 1)

当x = a - 1,v = a - (a - 1) + 1 = 2,当x = a,v = a - a + 1 = 1

= e^(- a - 1) · ∫(2→1) e^v/v dv

= e^(- a - 1) · ∫(2→1) e^x/x dx

= e^(- a - 1) · (- e) · (1/e)∫(1→2) e^x/x dx

= - e^(- a - 1 + 1) · A

= - e^(- a) · A

= - A/e^a

实际上,

∫(0→1) e^x/(1 + x) dx = [Ei(2) - Ei(1)]/e

∫(a - 1→a) e^(- x)/(x - a - 1) dx = e^(- a - 1) · (Ei(1) - Ei(2))

= e^(- a - 1) · - [Ei(2) - Ei(1)]/e · e

= e^(- a - 1) · - A · e

= - A/e^a