I am trying to find the interval for the value $p$ where the following integral converges (log is the natural logarithm):

$$\int_1^{\infty} \frac{\log{x}}{x^p} \,dx$$

So far I've used integration by parts to reduce the integral to the limit:

$\displaystyle \lim_{a\to\infty} \frac{\log{x} - p \log{x} - 1}{x^{p-1}(1-p)^2}$.

My intuition tells me that $p > 2$, but I am struggling to find a good argument.

A push in the right direction would be appreciated!

  • $\begingroup$ I think p>1 because any power of $x$ goes faster to $\infty$ then $\log x$ $\endgroup$
    – bob
    Jan 26, 2016 at 14:46

2 Answers 2


Recall that the logarithm function satisfies the inequality

$$\log(x)\le x-1<x \tag 1$$

for $x>0$.

Then, since $\log(x^{\alpha})=\alpha \log(x)$, we have from $(1)$ for $\alpha>0$

$$\begin{align} \alpha \log(x)&< x^{\alpha}\\\\ \log(x)&<\frac{x^{\alpha}}{\alpha} \\\\ \frac{\log(x)}{x^p}&< \frac{x^{\alpha-p}}{\alpha} \end{align}$$

Therefore, for any $p>1$ we can take $\alpha>0$ small enough so that $p-\alpha>1$. Therefore the integral of interest converges for all $p>1$ by the comparison test.

Since $\log(x)>1$ for $x>e$, then $\frac{\log(x)}{x^p}>\frac{1}{x^p}$ for $x>e$ and the integral diverges for $p\le 1$.

We can evaluate the improper integral applying integration by parts with $u=\log(x)$ and $v=x^{1-p}/(1-p)$. Proceeding accordingly, we obtain

$$\begin{align} \lim_{L\to \infty}\int_1^L \frac{\log(x)}{x^p}\,dx&=\lim_{L\to \infty}\left(\left. \frac{x^{1-p}\log(x)}{1-p}\right|_{1}^{L}\right)+\frac{1}{p-1}\lim_{L\to \infty }\int_1^L\frac{1}{x^p}\,dx\\\\ &=\frac{1}{(p-1)^2} \end{align}$$


It's clear that $p$ must be positive. Use integral test as the integrand is monotone decreasing, $$ \sum_{n=1}^\infty{\ln n\over n^p} $$ The above series converges iff by Cauchy condensation test the series below converges $$ \sum_{n=1}^\infty{n2^n\over2^{np}} $$ which converges if $p>1$ by root test.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .