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My task is to determine wetherthe integral $\int_e^\infty{\frac{\ln(x)^\alpha}{x}\,dx}$ does exist or not in depencence of $\alpha\in \mathbb{R}$.

For that I wrote $b$ instead of $\infty$ and then calculated the integral using substitution, which should be $\frac{1}{\alpha+1}((\ln(b))^{\alpha+1}-1)$. Now for the improper integral I get $\lim\limits_{b->\infty}(\int_e^b(\frac{\ln(x)^\alpha}{x}dx))=\frac{1}{\alpha+1}\cdot(\lim\limits_{b->\infty}\ln(b)^{\alpha+1}-1)$

And because $\lim\limits_{c->\infty}\ln(c)=\infty$ and $\lim\limits_{c->\infty} c^{\alpha+1} = \left\{ \begin{array}{ll} \infty & \alpha\geq -1 \\ 1 & \alpha=-1 \\ 0 & \alpha<-1 \end{array} \right. $

So the limit only exists for $\alpha<-1$, is this right?

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  • $\begingroup$ This is the right result $\endgroup$ – Dr. Sonnhard Graubner Jan 12 at 13:20
  • $\begingroup$ Why do you consider in the end the lim of a power of $c $ instead that of $\ln c $? $\endgroup$ – user Jan 12 at 13:25
  • $\begingroup$ because ln(c) is infinity anyway $\endgroup$ – Yefexem Jan 12 at 13:28
  • $\begingroup$ Does: $$\int{\frac{(\ln x)^{\alpha}}{x}dx}=\frac{(\ln x)^{\alpha +1}}{\alpha+1}+C$$ help? (FYI: I used IBP) $\endgroup$ – Rhys Hughes Jan 12 at 13:38
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$$I=\int_e^\infty\frac{\ln^a(x)}{x}dx$$ $$u=\ln(x),\,dx=xdu$$ $$I=\int_1^\infty u^adu=\left[\frac{u^{a+1}}{a+1}\right]_1^\infty$$ and this is clearly divergent

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  • $\begingroup$ Except if $\alpha < -1$, I think. $\endgroup$ – Claude Leibovici Jan 12 at 15:39
  • $\begingroup$ @ClaudeLeibovici yes that sounds right $\endgroup$ – Henry Lee Jan 12 at 15:41

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