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Honestly, I am asked to think about $$\int_{0}^{1} x^m \ln^\alpha(x) dx$$ And I applied all methods I know. I doubt if this integral makes sense either. If it is replicate, plz inform me to omit the question soon. Thanks.

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    This is annoying: why the downvote?!2012-06-08
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    I don't think this can be evaluated in elementary terms unless I'm overlooking something.2012-06-08
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    You might note that $\ln(x) < 0$ for $0 < x < 1$, so you need to use complex numbers to make sense of $(\ln x)^x$.2012-06-08
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    Babak, may I ask why were you asked to think about this?2012-06-08
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    A more conventional request would be the integral $\int_0^1 x^n \ln x\,dx$.2012-06-08
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    If you do allow complex numbers, the integrals certainly "make sense", i.e. converge, for $m > -1$. Note that $\ln(x)^x = e^{x \ln(\ln x)}$2012-06-08
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    Thanks for all comments. Yes I know that integral GEdgar. I were asked this written in an diffrental Equation book by someone. I didn't see such this Integral before. Maybe it was recorded wrong. Sorry if I made you bad feelings.2012-06-08
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    Babak, was it *printed* in the book, if so which book is it? Otherwise I think you should just skip that comment.2012-06-08
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    @BabakSorouh If you change $\log(x)^x$ to $\log(x)^\alpha$, then Gamma comes in, otherwise I don't know.2012-06-09
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    Maybe, it is $\alpha$ and not $x$ in the exponent - they are pretty similar in certain fonts...2012-06-09
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    @AD.: Thank you Sir for the time. Thanks.2012-06-09
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    @BabakSorouh It might be good to leave the original post as it was written, and then put the update at the bottom.2012-06-09

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$$ \begin{align} \int_0^1x^m\,\log^\alpha(x)\,\mathrm{d}x &=\int_{-\infty}^0u^\alpha\,e^{(m+1)u}\,\mathrm{d}u\\ &=(-1)^\alpha(m+1)^{-\alpha-1}\int_0^{\infty}t^\alpha e^{-t}\mathrm{d}t\\ &=(-1)^\alpha(m+1)^{-\alpha-1}\Gamma(\alpha+1) \end{align} $$