Proving the convergence of this integral (without evaluation)

Question

The integral in question is $$\int_{2}^{\infty}\frac{1}{\sqrt{e^x}\ln(x)}dx$$ I know this converges but I have no idea how to prove it. I figured I could use the limit comparison test but struggle to find a good test function for it since I need to know how $\sqrt{e^x}$ and $\ln(x)$ behave near infinity.

Any help to get me started? Is it possible to use another convergence test for this?

Answer 1

For $x\to\infty$ you can write $$ \frac{1}{\sqrt{e^x}ln(x)}=\frac{1}{\sqrt{e^x}}\frac{1}{1+\frac{ln(x)}{\sqrt{e^x}}}=\frac{1}{\sqrt{e^x}}\biggl(1-\frac{ln(x)}{\sqrt{e^x}}+o\bigg(\frac{ln(x)}{\sqrt{e^x}}\bigg)\biggl) $$ where the last equality is the McLaurin series of order 1. At this point should be easy to study convergence.

Answer 2

Note that $\frac{1}{\sqrt{e^x}\log{x}} < \frac{2}{\sqrt{e^x}}$ on the interval $[2,\infty)$.

We can easily evaluate the latter integral using only one substitution to be:

$$\int_2^\infty\frac{2}{\sqrt{e^x}}dx = \frac{4}{e}$$

By comparison we know that the integral must converge.