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.
Evaluating $\int_{0}^{1} x^m \ln^\alpha(x) dx$
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calculus
integration
definite-integrals
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0@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
1 Answers
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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} $