test convergence of improper integrals 4

Question

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.

Answer 1

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.

Answer 2

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 $\delta$.

Answer 3

$$ \lim_{x \to 1^{+}} (x-1) \frac{\sqrt{x}}{\ln x} = \lim_{x \to 1^{+}} \frac{\sqrt{x} + (x-1) \frac{1}{2 \sqrt{x}}}{\frac{1}{x}} = 1 $$

The integrand behaves like $\frac{1}{x-1}$ near $x=1$ and thus $ \displaystyle\int_1^2{\sqrt x\over \log x}dx$ diverges.

Answer 4

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.