# Improper Integrals

Determine whether the following improper integral is convergent or divergent.

$$\int_1^{\infty} \text{sech}\, x \ln x \,dx$$

I think that I need to use integration by parts but the sechx is really stumping me.

Thanks!

If $x\ge 1$ then $$0\le\frac{\ln x}{\cosh x}=\frac{2\ln x}{e^x+e^{-x}}\le \frac{2\ln x}{e^x+0}=\frac{2\ln x}{e^x}\le 2xe^{-x}$$ and apply comparison test.
$$\text{sech}\,x\ln x = \frac{2\ln x}{e^x+e^{-x}}.$$
$\ln x$ can be bounded by $\sqrt{x}$ and $\dfrac{1}{e^x+e^{-x}} \le \dfrac{1}{e^x}.$ Do you see how these help?
Starting from Tetori'answer, you could even show that $$\int \frac{2\ln x}{e^x}dx=2 \left(\text{Ei}(-x)-e^{-x} \log (x)\right)$$ and so $$\int_1^\infty \frac{2\ln x}{e^x}dx=-2 \text{Ei}(-1)\simeq 0.438768$$ while, numerically,$$\int_1^{\infty} \text{sech}\, x \ln x \,dx \simeq 0.430498$$