Question about the limit comparison test (for improper integrals) and intuition

Question

I am asked to tell for which $p > 0$ values the integral $$\int_1^2 \frac {e^{-px}} {\ln(x)} \mathrm{d}x$$ converges. I tried to use the limit comparison test as follows:

$$\frac {\frac {e^{-px}} {\ln(x)}} {\frac 1 {\ln(x)}} = e^{-px} \underset {x \to 1^+} \to e^{-p} > 0$$

and since $\int_1^2 \frac {\mathrm{d}x} {\ln(x)}$ diverges so does the discussed integral.

Is this correct? It doesn't make sense to me that even if we put $p=100$ the $\frac 1 {\ln(x)}$ will make it diverge... In that case can anyone try to explain the intuition behind that?

Thanks :)

Answer 1

The function $e^{-px}$ is bounded below (and away from $0$) on the interval $[1,2]$. A direct comparison gives $$\int_1^2 \frac{e^{-px}}{\ln x} \, dx \ge e^{-2p} \int_1^2 \frac 1{\ln x} \, dx$$ so that the integral diverges.

Another way to look at it is no matter how large $p$ is, the fact that $e^{-px}$ is positive (regardless of how small) around $x=1$ means that it does not counterbalance the singularity of $\frac 1{\ln x}$.

Answer 2

$e^{-px}$ is decreasing, so on the interval of integration it is bounded between $e^{-p}$ and $e^{-2p}$. Hence for any $p>0$, $$ \int_1^2 \frac{e^{-p}}{\log{x}} \, dx \geq \int_1^2 \frac{e^{-px}}{\log{x}} \, dx \geq \int_1^2 \frac{e^{-2p}}{\log{x}} \, dx, $$ so the convergence or divergence of the integral is completely determined by that of $\int_1^2 dx/\log{x}$.