Question

第二题这个反常积分最小值怎么求

Answer 1

∫(2->+∞) dx/[x(lnx)^k]
=∫(2->+∞) dlnx/(lnx)^k
= [1/(-k+1)] [ (lnx)^(-k+1)]|(2->+∞)
收敛
=>-k+1<0
=>k>1
∫(2->+∞) dx/[x(lnx)^k]
=∫(2->+∞) dlnx/(lnx)^k
= [1/(-k+1)] [ (lnx)^(-k+1)]|(2->+∞)
=-[1/(-k+1)] (ln2)^(-k+1)
let
f(x) =-[1/(-x+1)] (ln2)^(-x+1)
= [1/(x-1) ](ln2)^(1-x)
f'(x)
= [-1/(x-1)^2 - ln(ln2)/(x-1) ](ln2)^(1-x)
f'(x) = 0
[-1/(x-1)^2 - ln(ln2)/(x-1) ](ln2)^(1-x) =0
-1 -(x-1) lnln2 =0
x-1= -1/ln(ln2)
x = 1- 1/ln(ln2)
f'(x) | x= (1- 1/ln(ln2))+ >0
f'(x) | x= (1- 1/ln(ln2))- <0
x=1- 1/ln(ln2) ( min)
min ∫(2->+∞) dx/[x(lnx)^k] at k = 1- 1/ln(ln2)

追问

为什么最小值在发散的时候求呢

追答

最小值 ∫(2->+∞) dx/[x(lnx)^k]
=最小值 [1/(x-1) ](ln2)^(1-x)
f'(x0)=0 , f'(x0+)>0 and f'(x0-) min f(x) =f(x0)
以下是计算 min f(x) 的步骤
f(x) =-[1/(-x+1)] (ln2)^(-x+1)
= [1/(x-1) ](ln2)^(1-x)
f'(x)
= [-1/(x-1)^2 - ln(ln2)/(x-1) ](ln2)^(1-x)
f'(x) = 0
[-1/(x-1)^2 - ln(ln2)/(x-1) ](ln2)^(1-x) =0
-1 -(x-1) lnln2 =0
x-1= -1/ln(ln2)
x = 1- 1/ln(ln2)
f'(x) | x= (1- 1/ln(ln2))+ >0
f'(x) | x= (1- 1/ln(ln2))- <0
x=1- 1/ln(ln2) ( min)

追问

我懂了，谢谢你哟

😀😃