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For what values of $\alpha$ and $\beta$ the following improper integral converge? $$ \int_{1}^{\infty}\frac{(\ln x)^{\alpha}}{x^{\beta}}\,dx $$

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closed as off-topic by heropup, M. Vinay, choco_addicted, Shailesh, user91500 Jun 16 '16 at 7:18

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  • $\begingroup$ Just off the top of my head, have you tried $u = \ln x \implies (\ln x)^{\alpha}/x^{\beta} = u^\alpha e^{1-\beta}$. Then there should exists some recursive relation for partial integration. $\endgroup$ – user305860 Jun 16 '16 at 5:43
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Let $I(\alpha ,\beta)$ be define by

$$I(\alpha,\beta)=\int_1^\infty \frac{\log^\alpha(x)}{x^\beta}\,dx $$

Enforce the substitution $x\to e^x$ to obtain

$$I(\alpha,\beta)=\int_0^\infty u^\alpha e^{(1-\beta)u}\,du \tag 1$$

The integral in $(1)$ is improper at both upper and lower limits. At the upper limit, we require $1-\beta<0$ for convergence while at the lower limit, we require $\alpha >-1$ for convergence.

The integral converges for $\alpha>-1$ and $\beta>1$

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