How do you prove that ∫101√ln(1x)dx converges? I've tried more or less everything I can think of and still can't get the answer. Any hints will be appreciated!
Answer
HINT
We have that
∫101√ln(1x)dx=∫∞11x2√lnxdx=∫211x2√lnxdx+∫∞21x2√lnxdx
and then refer to limit comparison test.
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