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Express this integral through Euler's integral. $$\int_{0}^{1}\left(\ln \frac1x\right)^pdx, \ \ \ (p>-1)$$

I tried taking $u=\ln\frac1x$, then $du=-\frac1x$ but it doesn't seem to help much. Any ideas how to solve this problem?

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$$\ln\frac1x=\ln(x)^{-1}=-\ln x$$ and set $$\ln x=u$$ –  lab bhattacharjee May 25 at 12:05
    
@labbhattacharjee You got your exponent in the wrong place there. –  Cole Johnson May 25 at 21:23

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up vote 9 down vote accepted

Let $x=e^{-u}, dx=-e^{-u}du$ then it follows that: $$ \begin{aligned} \int _{0}^{1}\! \left( \ln \left( \frac{1}{x} \right) \right) ^{p}{dx}=& \int _{0}^{\infty }\!{(\ln(e^{u}))}^{p}{{\rm e}^{-u}}{du}\\ =&\int _{0}^{\infty }\!{u}^{p}{{\rm e}^{-u}}{du}\\ =&\Gamma \left( p+1 \right) \end{aligned}$$

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If $p\in \mathbb{Z}$ then it equals $p!$. Thanks for the great answer. –  RasmusE May 25 at 12:20

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