Question

高数题目 反常积分敛散性

答案是D 但我感觉B也是对的 求解答

Answer 1

I = ∫<1, +∞>dx/[x√(2x^2-2x+1) = ∫<1, +∞>dx/[x√[2(x-1/2)^2+1/2]
= (1/√2)∫<1, +∞>dx/[x√[(x-1/2)^2+1/4]
令 x - 1/2 = (1/2)tant, 则 √[(x-1/2)^2+1/4] = (1/2)sect,
x = (1/2)(1+tant), dx = (1/2)(sect)^2 dt
I = (1/√2)∫<π/4, π/2>(1/2)(sect)^2 dt/[(1/2)(1+tant)(1/2)sect]
= √2 ∫<π/4, π/2>sectdt/(1+tant) = √2 ∫<π/4, π/2>dt/(cost+sint)
= ∫<π/4, π/2>dt/cos(t-π/4) = ∫<π/4, π/2>sec(t-π/4)d(t-π/4)
= {ln[sec(t-π/4)+tan(t-π/4)]}<π/4, π/2>
= ln[sec(π/4)+tan(π/4)] - ln[sec0+tan0]
= ln(1+√2) - ln1 = ln(1+√2)

Answer 2

好像B也是对的，相当于∫(0，1)1/√(1-x)dx,是收敛的。

Answer 3

I = (1/√2)∫(1/2)(sect)^2 dt/[(1/2)(1+tant)(1/2)sect]