I am looking to study the convergence of this integral $$ \int_0^1 \frac{x}{\ln x}$$

One method to prove the convergence is to use the special case of Cauchy definition "what is called in French l'integral de Bertrand" as

The integral $\int_0^{1/e} \frac{1}{x^a (\ln x)^b}$ converges iff $(a <1)$ or ($a=1$ and $b >1$)

so in my question $a=-1$ and the integral converges but I am looking for another method to do so. Is there another simple way to prove the convergence.

  • $\begingroup$ Your integral actually diverges because of the singularity of the integrand around $x=1$. $\endgroup$ Nov 4, 2019 at 8:07
  • $\begingroup$ Your inequality is not true, the integrand in the LHS is positive and the RHS is negative. $\endgroup$
    – user
    Nov 4, 2019 at 8:19
  • $\begingroup$ @user Yes you are right...I see now $\endgroup$ Nov 4, 2019 at 8:20
  • $\begingroup$ Your original integral converges, but this is because the singularity of the denominator $\ln x$ at $x=1$ is cancelled by the numerator $x-1$. You definitely have to take this effect into consideration. $\endgroup$ Nov 4, 2019 at 8:27

2 Answers 2


We have that

$$\int_0^1 \frac{x}{\ln x}dx=\int_1^\frac12 \frac{x}{\ln x}dx+\int_\frac12^1 \frac{x}{\ln x}dx$$

and since $\lim_{x\to 0^+} \frac{x}{\ln x}=0$ all boils down in $\int_\frac12^1 \frac{x}{\ln x}dx$ and by $x=1-y$ we obtain

$$\int_\frac12^1 \frac{x}{\ln x}dx=\int_0^\frac12 \frac{1-y}{\ln (1-y)}dy$$

which diverges by limit comparison test with $\int_0^\frac12 \frac1ydy$ indeed

$$\lim_{x\to 0^+} \frac{\frac{1-y}{\ln (1-y)}}{\frac1y}=\lim_{x\to 0^+} (1-y)\frac{y}{\ln (1-y)}=1$$

In a similar way we can prove that for $b>1$

$$\int_0^1 \frac {x} { (\ln x)^b }dx$$

the integral converges.

  • $\begingroup$ thanks a lot. I added an edit as you see above $\endgroup$ Nov 4, 2019 at 8:13
  • $\begingroup$ Yes I understand now and as you said from the beginning my first inequality not correct. Could you please clarify why $\int_0^1 \frac{x-1}{\ln x}$ converges? $\endgroup$ Nov 4, 2019 at 9:02
  • $\begingroup$ I know that $\ln (1+x) \sim x$ as $x\rightarrow 0$ and $\ln (x) \sim x-1$ as $x\rightarrow 1$ but how does this enough to show the convergence? $\endgroup$ Nov 4, 2019 at 9:05
  • 1
    $\begingroup$ We have that $\lim_{x\to 0^+}\frac{x-1}{\ln x}=0$ and $\lim_{x\to 1^-}\frac{x-1}{\ln x}=1$ therefore it is a proper integral which converges. $\endgroup$
    – user
    Nov 4, 2019 at 9:07

Are you sure that the integral converges ? The integrand is equivalent to $\frac1{x - 1}$ around $1$.

  • $\begingroup$ thanks for your interest . I added an edit $\endgroup$ Nov 4, 2019 at 8:14
  • $\begingroup$ The observation is fine, maybe you should explain better in which sense it is equivalent. $\endgroup$
    – user
    Nov 4, 2019 at 8:34

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