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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?! –  DonAntonio Jun 8 '12 at 20:21
I don't think this can be evaluated in elementary terms unless I'm overlooking something. –  Joe Jun 8 '12 at 20:22
You might note that $\ln(x) < 0$ for $0 < x < 1$, so you need to use complex numbers to make sense of $(\ln x)^x$. –  Robert Israel Jun 8 '12 at 20:22
Babak, may I ask why were you asked to think about this? –  AD. Jun 8 '12 at 20:25
A more conventional request would be the integral $\int_0^1 x^n \ln x\,dx$. –  GEdgar Jun 8 '12 at 20:25

1 Answer 1

up vote 4 down vote accepted

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

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+1 That was what I meant. –  AD. Jun 9 '12 at 11:53

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