test convergence of improper integrals 4 Test the convergence of improper integrals :
$$\int_1^2{\sqrt x\over \log x}dx$$
I basically have no idea how to approach a problem in which log appears. Need some hint on solving this type of problems.
 A: Consider the following:
$$\lim_{\epsilon \to 0} \int_{\epsilon}^1 dx \frac{\sqrt{1+x}}{\log{(1+x)}}$$
Now, for the bottom limit of the integral, note that
$$\log{(1+\epsilon)} \sim \epsilon$$
so that, near this limit, the integrand behaves as $1/\epsilon$ as $\epsilon \to 0$, or as $1/x$ as $x \to 0$.  This represents a non-integrable singularity (the integral would behave as $\log{\epsilon}$ near this limit), and therefore the integral diverges.
A: Let, $I(\delta)=\displaystyle\int_{1+ \delta}^2{\sqrt x\over \log x}dx$
$$I(\delta) \ge \displaystyle\int_{1+ \delta}^2{1\over \log x} \geq \displaystyle\int_{1+ \delta}^2{1\over x-1}=-\log\delta$$
As, $\log x \le x-1 $ for $x\ge 1$.
Letting $\delta \to 0$ you can have that your integral diverges, as $I(\delta)$ is continuousin $\delta$.
A: $$ \lim_{x \to 1^{+}} (x-1) \frac{\sqrt{x}}{\ln x} = \lim_{x \to 1^{+}} \frac{\sqrt{x} + (x-1) \frac{1}{2 \sqrt{x}}}{\frac{1}{x}} = 1 $$
The integrand behaves like $\frac{1}{x-1}$ near $x=1$ and thus $ \displaystyle\int_1^2{\sqrt x\over \log x}dx$ diverges.
A: First make the change of variables $\ln(x)=u$
$$\int_1^2{\sqrt x\over \log x}dx =\int _{0}^{\ln  \left( 2 \right) }\!{\frac {{{\rm e}^{3/2\,u}}}{u}}{du}.$$
Now, you can see that the integrand behaves as
$$ {\frac {{{\rm e}^{3/2\,u}}}{u}}\sim _{u\to 0} \frac{1}{u} $$
which is not integrable on the given interval. 
