Does $\int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}}$ converge?

Question

I have improper integral $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}}$$ and I need to find out if it converges/diverges. Now I don't if my thinking is correct:

$\displaystyle \int \frac{dx}{\sqrt{\ln(x)}}$ goes to $0$ slower than $\displaystyle \int \frac{dx}{\sqrt{x}}$, thus from integral comparison: $$\displaystyle \int_{1}^{2}\frac{dx}{\sqrt{(2-x)\ln(x)}} < \int_{1}^{2} \frac{dx}{\sqrt{x}}$$ it diverges. Is this correct? Thanks.

Answer 1

Bounding above the integral by a convergent one does not prove the integral diverges.

Actually it converges. The integrand is non-negative and

• at $2$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{(2-x)\log 2}$, and the integral $\int_{3/2}^{2}\frac{\mathrm dx}{\sqrt{2-x}}$ is convergent;
• at $1$, using $\log(1+t)\overset{t\to 0}{\sim}t$, we have $\frac 1{(2-x)\log (x)}\sim \frac 1{\log(x-1+1)}\sim \frac 1{x-1}$.

Answer 2

I solve the improper integral firstly by take it apart at $3/2$:

$$\int_1^{3/2}f(x)dx+\int_{3/2}^2f(x)dx$$

Now in the first integral we see that $$\lim_{x\to 1^+}\sqrt{x-1}f(x)=1<\infty$$ And for the second one we get $$\lim_{x\to 2^-}\sqrt{2-x}f(x)=1/\sqrt{\ln 2}<\infty$$ Therefore, both are converge and so the summation converges.