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Math Help - limit with integral

  1. #1
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    limit with integral

    Find:
    \lim_{r \rightarrow 0^+} \frac{1}{r|\ln r|^{\alpha}} \int^{r}_{0} |\ln x|^{\alpha} e^{-x^2} dx for \alpha \in (0;1).
    My intuition says that would be 0 so I have tried to use squeeze theorem. This limit is bounded below by 0.
    I've thought I could bound it above using Hölder's inequality with p=\frac{1}{\alpha} and q=\frac{1}{1-\alpha} now what I have got:
     \frac{1}{r|\ln r|^{\alpha}} \int^{r}_{0} |\ln x|^{\alpha} e^{-x^2} dx \leq \left( \frac{\int^{r}_{0} |\ln x|dx}{|\ln x| \int^{r}_{0} e^{\frac{-x^2}{1-\alpha}}dx} \right)^{\alpha} \cdot \frac{\int^{r}_{0}e^{\frac{-x^2}{1-\alpha}}dx}{r} . Does it get me anywhere?
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  2. #2
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    Re: limit with integral

    I think I solved the problem:
    limit with integral.pdf.
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