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

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)]dx 志在羊城 1年前他留下的回答 已收到2个回答

panjian8264 春芽

该名网友总共回答了21个问题，此问答他的回答如下：采纳率：81%

∫(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

1年前他留下的回答

5

harryabc12 网友

该名网友总共回答了150个问题，此问答他的回答如下：

∫[a-1,a] [(e^(-x)/(x-a-1)]dx
令x=a-u，则dx=-du，x：a-1--->a，u：1--->0
=-∫[1,0] [(e^(u-a)/(-u-1)]du
=-∫[0,1] [(e^(u-a)/(u+1)]du
=-e^(-a)∫[0,1] [(e^u/(u+1)]du
=-e^(-a)A

1年前他留下的回答

0

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