how to show $\int_{1}^{\infty}\frac{(\log x)^a}{x^p}dx$, if $p \gt 1$ and $a \gt 0$???. I can show this is not improper integrable in case $p \lt 1 $, but I was stuck in showing the other case. Could you let me know how to handle it??
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It is not integrable if $p<1$. – Bernard Apr 30 '20 at 10:27
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@Bernard Yes, I was wrong to post. I will edit! Thanks, but could you let me know how to show this integral is finite if $p >1$? – fivestar Apr 30 '20 at 10:29
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You can show that the integrand is dominated by $1/x^{p-\delta}$ for arbitrarily small $ \delta > 0 $ and then use the fact that $1/x^{p-\delta}$ is integrable for sufficiently small $\delta$ (provided $p>1$). – derpy Apr 30 '20 at 10:31
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@derpy Actually, I tried to show that you mentioned but, I can't... – fivestar Apr 30 '20 at 10:35
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$$ x\mapsto e^u $$ – Jack D'Aurizio Apr 30 '20 at 10:46
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Do some asymptotic analysis:
Choose $\varepsilon>0$ such that $1<q=p-\varepsilon$, and rewrite the integrand as $$\frac{\log^ax}{x^p}=\frac{\log^ax}{x^\varepsilon}\frac 1{x^{q}}=o\Bigl(\frac1{x^q}\Bigr),$$ and the latter function has a convergent integral on $[1,\infty)$ since $q>1$.
Bernard
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I understand $\frac{1}{x^q}$ is convergent, but how do I think $\frac{\log^ax}{x^\varepsilon}$ ?? – fivestar Apr 30 '20 at 11:17
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Using the comparison test: as the ratio tends to $0$ at $\infty$, it is $<1$ if $x$ is large enough, so the function is bounded from above by $\frac 1{x^q}$. (Actually, it is a basic result in asymptotic analysis). – Bernard Apr 30 '20 at 11:36