Convergence of integral $\int_1^2 \frac{\sqrt{x}} {\ln(x)} \,dx $

Question

I want to determine whether or not the integral $$\int_1^2 \frac{\sqrt{x}} {\ln(x)} \,dx $$ converges. I have tried things like $$\int_1^2 \frac{\sqrt{x}} {\ln(x)} \,dx \leq \int_1^2 \frac{2} {\ln(x)} dx ,$$ but I find myself unable to evaluate the latter integral. Next I try: $$\int_1^2 \frac{\sqrt{x}} {\ln(x)} dx \geq \int_1^2 \frac{\sqrt{x}} x dx .$$ In this case the latter integral is finite but that does not tell me anything about the convergence or divergence of the original integral. What comparison can I make to determine the convergence of this integral?

Answer 1

For a direct comparison, since $\ln x < x-1$ for $x > 1$, we have

$$\frac{\sqrt{x}}{\ln x} > \frac{\sqrt{x}}{x -1} > \frac{1}{x -1} $$

Note that $\displaystyle\int_1^2 \frac{dx}{x-1}= -\lim_{c \to 1+} \ln(c-1)=+\infty.$

Answer 2

Instead of looking for comparison function $f(x)$, just look at how the denominator, $\ln{x}$, approaches $0$ near $x=1$. Therefore, $f(x)$ approaches $\infty$ near $x=1$ and so the integral diverges.

Here is a visual representation of $\int_1^2 \frac{\sqrt{x}}{\ln{x}} \; dx$:

Answer 3

Put $x=e^z,$ $dx=e^z dz$: $$ \int _1^2 \frac{\sqrt{x}}{\ln x}\,dx = \int_0 ^{\ln 2} \frac{e^{3z/2}}{z}\,dz \geq \int_0 ^{\ln 2} \frac{1}{z}\,dz = \infty $$