I'm working in this problem and I'm having some problems.
Study the convergence of this improper integral:
∫120dtta|ln(t)|b,a,b>0
For a<1 I've compared it with the integral
∫120dtta
and found that is convergent. When a=0, taking u=ln(t) and du=dtt we have:
∫120dtt|ln(t)|b=∫ln(12)−∞duub
which is convergent for b>1 and divergent for b≤1 (is this correct?).
When a>1 I think that diverges, but cannot prove it. Any hint?
Answer
Set u=−logt. Then t=e−u, so dt=−e−udu. The limits become ∞ and log2, and we have
∫∞log2u−be−(1−a)udu
Now, the integrand is bounded, so the problem is only for large u. In particular, there are now several cases to examine:
- a>1. The integral diverges, because eku grows faster than u−b shrinks for any k>0, so the integrand does not tend to zero as u→∞.
- a=1. The integral is ∫∞log2u−bdu, which we know converges if and only if b>1.
- $00,because∫∞log2u−be−(1−a)udu<∫∞log2(log2)−be−(1−a)udu=(log2)−b11−a.$
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