As its told on the title I want to check the convergence/divergence of the improper integral when a∈R: ∫10∣log(x)∣a√1−x2dx
So, it's improper at x=0 and x=1, so I split the integral in :
∫120∣log(x)∣a√1−x2dx+∫112∣log(x)∣a√1−x2dx
I see, that on the first one the integrals it's like ∫∣log(x)∣adx by the comparison limit test, but I don't know how to prove that ∫∣log(x)∣adx converges.
Hopefully you can help me. Much thanks!
Answer
Note that
∫120∣log(x)∣a√1−x2dx
converges ∀a by comparison test with 1√x.
For the second part
∫112∣log(x)∣a√1−x2dx
let 1−x2=y2
∫0√32∣log(√1−y2∣ay⋅(−y√1−y2)dy=∫√320∣log(√1−y2∣a√1−y2dy
and note that for y→0
∣log(√1−y2)∣∼y22
thus the integral converges for −2a<1 that is a>−12 and diverges for −2a≥1 that is a≤−12 by comparison with y2a.
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