Does the following expression converge? Where $n$ is positive integer $1,2,3,...$
$$\int_0^\infty(\ln x)^n dx$$
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$\begingroup$ Diverges for even n. $\endgroup$– Edwin_RJun 10, 2014 at 9:59
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$\begingroup$ @Edwin_R, how do you propose I go about evaluating it? $\endgroup$– SA-255525Jun 10, 2014 at 10:03
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$\begingroup$ Why don't we just compare it with the integral of one from $e$ to $\infty$? $\endgroup$– JP McCarthyJun 10, 2014 at 10:53
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$\begingroup$ @Edwin_R Diverges for every $n$ I would claim... $\endgroup$– ThorbenJun 11, 2014 at 10:08
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$\begingroup$ @Thorben yes. and now i know, after reading your answer :-). i wrote that because it was obvious for even n. $\endgroup$– Edwin_RJun 11, 2014 at 19:36
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3 Answers
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I hope this is correct,
A substitution $y=ln(x)$ gives, $$\int_0^{\infty}\ln(x)^ndx=\int_{-\infty}^{\infty}y^ne^ydy=\int_{-\infty}^0y^ne^ydy+\int_0^{\infty}y^n e^ydy $$ where $$\int_{-\infty}^0y^ne^ydy=\int^{\infty}_0(-y)^ne^{-y}dy=(-1)^n\int^{\infty}_0y^ne^{-y}dy=(-1)^nn!$$ but $$ \int_0^{\infty}y^n e^ydy\geq\int_0^{\infty}y^ndy=\lim_{y\rightarrow \infty}\frac{1}{n+1}y^{n+1}=\infty$$ Hence $\int_0^{\infty}\ln(x)^ndx$ is divergent for every $n\geq0$.
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$\begingroup$ how can limits after substitution be same?In the first step. $\endgroup$ Jun 11, 2014 at 10:27
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$\begingroup$ @SA-255525 where do you mean? After substitution you integrate from $-\infty$ to $\infty$. Sorry I just see that i forget to put the minus in front of the lower limit... $\endgroup$– ThorbenJun 11, 2014 at 10:28
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$\begingroup$ @SA-255525 Yes, just plot $ln(x)$ somewhere and you will see what happends if $x$ tends to zero $\endgroup$– ThorbenJun 11, 2014 at 10:32
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$\begingroup$ Thanks very much I missed this point totally. And yes this integral is divergent for every n>=0. your help is much appreciated. $\endgroup$ Jun 11, 2014 at 10:35
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You can use integration by parts to reduce the exponent step by step. That will lead to a formula like this. $$\int(\ln x)^n =x(\ln x)^n-n\int(\ln x)^{n-1}dx$$
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Split the interval $(0,\infty)$ into $(0,1)$ and $(1,\infty)$. For the former, see $\Gamma$ function. As for the latter, it clearly diverges, thus making the entire quantity divergent.
