# Evaluate logrithmic integral

Does the following expression converge? Where $n$ is positive integer $1,2,3,...$

$$\int_0^\infty(\ln x)^n dx$$

• Diverges for even n. Jun 10, 2014 at 9:59
• @Edwin_R, how do you propose I go about evaluating it? Jun 10, 2014 at 10:03
• Why don't we just compare it with the integral of one from $e$ to $\infty$? Jun 10, 2014 at 10:53
• @Edwin_R Diverges for every $n$ I would claim... Jun 11, 2014 at 10:08
• @Thorben yes. and now i know, after reading your answer :-). i wrote that because it was obvious for even n. Jun 11, 2014 at 19:36

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 $n\geq0$.

• how can limits after substitution be same?In the first step. Jun 11, 2014 at 10:27
• @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... Jun 11, 2014 at 10:28
• Is ln 0 = -inf. Jun 11, 2014 at 10:31
• @SA-255525 Yes, just plot $ln(x)$ somewhere and you will see what happends if $x$ tends to zero Jun 11, 2014 at 10:32
• Thanks very much I missed this point totally. And yes this integral is divergent for every n>=0. your help is much appreciated. Jun 11, 2014 at 10:35

You can use integration by parts to reduce the exponent step by step. That will lead to a formula like this. $$\int(\ln x)^n =x(\ln x)^n-n\int(\ln x)^{n-1}dx$$

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.