Improper integral of $\int_1^2 \frac{\sqrt{x}}{\log(x)}\,\mathrm dx$

Question

The question is to check the convergence of the improper integral of $\int_1^2 \frac{\sqrt{x}}{\log(x)}\,\mathrm dx$. I tried using comparision test and tried to convert or compare it to the Gamma function but it led to nowhere .

Answer 1

$$\int_1^2 \frac{\sqrt{x}}{\log(x)}\,\mathrm dx\geq\int_1^2 \frac{1}{x\log(x)}\,\mathrm dx\to+\infty$$

Answer 2

We compare the integrand with the function $1/(x-1)$. Note that $$\lim_{x\to 1^+}\frac{\sqrt{x}(x-1)}{\ln(x)}=\lim_{t\to 0^+}\frac{t}{\ln(1+t)}=1$$ which implies that there is $r\in (1,2]$ such that for all $x\in (1,r]$ $$\frac{\sqrt{x}(x-1)}{\ln(x)}\geq \frac{1}{2}\implies \frac{\sqrt{x}}{\ln(x)}\geq \frac{1}{2(x-1)}.$$ Therefore $$\int_1^2\frac{\sqrt{x}}{\ln(x)}\, dx\geq \frac{1}{2}\int_1^r\frac{1}{x-1}\,d x=\frac{1}{2}\left(\ln(r-1)-\lim_{x\to 1^+}\ln(x-1)\right)=+\infty.$$