问一道高数题,烦请大神解答?
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∫(2->+∞) dx/[x(lnx)^k]
=∫(2->+∞) dlnx/(lnx)^k
=-[1/(k-1)] [ 1/(lnx)^(k-1) ] |(2->+∞)
收敛
k-1>0
k>1
k>1
∫(2->+∞) dx/[x(lnx)^k]
=-[1/(k-1)] [ 1/(lnx)^(k-1) ] |(2->+∞)
=[1/(k-1)](ln2)^(k-1)
let
k>1
f(k) = [1/(k-1)](ln2)^(k-1)
f'(k)= [-1/(k-1)^2](ln2)^(k-1) +(ln2)^(k-2) . ln(ln2)
f'(k) =0
[-1/(k-1)^2](ln2)^(k-1) +(ln2)^(k-2) . ln(ln2) =0
-ln2 +(k-1)^2 .ln(ln2) =0
(k-1)^2 = ln2/ln(ln2)
k= 1+ √[ln2/ln(ln2)]
min f(x)
=f(1+ √[ln2/ln(ln2)] )
=[1/√[ln2/ln(ln2)] ]. (ln2)^(√[ln2/ln(ln2)])
=∫(2->+∞) dlnx/(lnx)^k
=-[1/(k-1)] [ 1/(lnx)^(k-1) ] |(2->+∞)
收敛
k-1>0
k>1
k>1
∫(2->+∞) dx/[x(lnx)^k]
=-[1/(k-1)] [ 1/(lnx)^(k-1) ] |(2->+∞)
=[1/(k-1)](ln2)^(k-1)
let
k>1
f(k) = [1/(k-1)](ln2)^(k-1)
f'(k)= [-1/(k-1)^2](ln2)^(k-1) +(ln2)^(k-2) . ln(ln2)
f'(k) =0
[-1/(k-1)^2](ln2)^(k-1) +(ln2)^(k-2) . ln(ln2) =0
-ln2 +(k-1)^2 .ln(ln2) =0
(k-1)^2 = ln2/ln(ln2)
k= 1+ √[ln2/ln(ln2)]
min f(x)
=f(1+ √[ln2/ln(ln2)] )
=[1/√[ln2/ln(ln2)] ]. (ln2)^(√[ln2/ln(ln2)])
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