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I'm working in this problem and I'm having some problems.

Study the convergence of this improper integral:

$$\displaystyle\int_0^{\frac12}\dfrac{\mathrm{d}t}{t^a \lvert\ln(t)\rvert^b},\quad a,b>0$$

For $\boxed{a<1}$ I've compared it with the integral $$\int_0^{\frac12}\dfrac{\mathrm{d}t}{t^a}$$ and found that is convergent. When $\boxed{a=0}$, taking $u=\ln(t)$ and $du=\frac{\mathrm{d}t}{t}$ we have: $$\int_{0}^{\frac12}\dfrac{\mathrm{d}t}{t\lvert\ln(t)\rvert^b}=\int_{-\infty}^{\ln(\frac12)}\dfrac{\mathrm{d}u}{u^b}$$ which is convergent for $b>1$ and divergent for $b\leq 1$ (is this correct?). When $\boxed{a>1}$ I think that diverges, but cannot prove it. Any hint?

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  • $\begingroup$ Have you formulated the required improper integral in terms of limits? The reason these integrals are improper is that the expression is unbounded at $t=0$. $\endgroup$ Commented Jun 6, 2015 at 21:34

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Set $u=-\log{t}$. Then $t = e^{-u} $, so $dt = -e^{-u} \, du$. The limits become $\infty$ and $\log{2}$, and we have $$ \int_{\log{2}}^{\infty} u^{-b} e^{-(1-a)u} \, du $$ Now, the integrand is bounded, so the problem is only for large $u$. In particular, there are now several cases to examine:

  1. $a>1$. The integral diverges, because $e^{ku}$ grows faster than $u^{-b}$ shrinks for any $k>0$, so the integrand does not tend to zero as $u \to \infty$.
  2. $a=1$. The integral is $\int_{\log{2}}^{\infty} u^{-b} \, du$, which we know converges if and only if $b>1$.
  3. $0<a<1$ The integral converges for any $b>0$, because $$ \int_{\log{2}}^{\infty} u^{-b} e^{-(1-a)u} \, du < \int_{\log{2}}^{\infty} (\log{2})^{-b} e^{-(1-a)u} \, du = (\log{2})^{-b} \frac{1}{1-a}. $$
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  • $\begingroup$ That's exactly what I was looking for. Thank you very much for your help! $\endgroup$
    – user246336
    Commented Jun 6, 2015 at 21:50

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