# The convergence of an integral

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.

• Your integral actually diverges because of the singularity of the integrand around $x=1$. Nov 4, 2019 at 8:07
• Your inequality is not true, the integrand in the LHS is positive and the RHS is negative.
– user
Nov 4, 2019 at 8:19
• @user Yes you are right...I see now Nov 4, 2019 at 8:20
• 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. Nov 4, 2019 at 8:27

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.

• thanks a lot. I added an edit as you see above Nov 4, 2019 at 8:13
• 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? Nov 4, 2019 at 9:02
• 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? Nov 4, 2019 at 9:05
• 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.
– user
Nov 4, 2019 at 9:07

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

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