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Question #70731

1 Answer

Feb 20, 2017

See below.

Explanation:

#logx < x->1/(logx sqrtx) > 1/(x sqrtx)#

but

#int_a^b (dx)/(x sqrtx) = 2/sqrt(a)-2/sqrt(b) = lim_(a->0) 2/sqrt(a)-2/sqrt(1) = oo#

Then if #int_a^b (dx)/(x sqrtx) < int_a^b (dx)/(logx sqrtx) # we have also that

#lim_(a->0)int_a^1(dx)/(logx sqrtx) # is divergent

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