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Let $n \in \mathbb N$ be arbitrary. Does the integral
$$\int_{0}^{1} |\ln (x)|^n \, dx$$ converge? I asked myself this question and I have no idea of a proof or counter example. Someone can give me a help?

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  • $\begingroup$ This is Euler's first integral expression for the $\Gamma$ function. It evaluates to $n!$ for $n>-1$. $\endgroup$ – Lucian Dec 21 '14 at 17:44
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The only problem is at $0$. The absolute value is irrelevant (just multiply by $(-1)^{n+1}$ the final result when no absolute value is used), so we can consider $$ \int_a^1(\log x)^n\,dx $$ with $a>0$. An integration by parts gives $$ \Bigl[x(\log x)^n\Bigr]_a^1-\int_a^1 x\cdot n(\log x)^{n-1}\frac{1}{x}\,dx $$ Can you go on from here?

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For $n$ even and $a>0$: $\int_{a}^{1} |lnx|^n dx=\int_{a}^{1}(x)^{'}(lnx)^ndx=xlnx_{a}^{1}-\int_{a}^{1}xn(lnx)^{n-1}\frac {1}{x}dx=xlnx_{a}^{1}-\int_{a}^{1}n(lnx)^{n-1}dx$. By induction $\int_{a}^{1} (lnx)^{n}dx$ converges iff $\int_{a}^{1} lnx dx$ converges.

Same tactic for $n$ odd.

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