Convergence of an improper integral $ \log( 1 + 2\operatorname{sech}x)$

Question

Show that the following integral is convergent

$ \int_{0}^{∞} \log (1+2\operatorname{sech} x) dx $

I tried by replacing $\operatorname{sech}x$ with $ 2/ e^x + e^{-x} $

and then by using limit comparison test with $ g(x) = 1/x^2 $ But couldn't solve further.

Answer 1

$$ 0\leq\int_{0}^{+\infty}\log\left(1+2\operatorname{sech}x\right)\,dx \leq 2\int_{0}^{+\infty}\operatorname{sech}(x)\,dx =\pi $$ since $0\leq \log(1+z)\leq z$ for any $z\geq 0$. Actually

$$ \int_{0}^{+\infty}\log\left(1+2\operatorname{sech}x\right)\,dx =\frac{\pi^2+4\operatorname{arccosh}(2)^2}{8}\approx 2.1008896. $$

Answer 2

Quite simple with equivalents: the positive function $$\operatorname{sech}x\sim_{+\infty}\frac 2{\mathrm{e}^x}$$ and the integral $\;\displaystyle\int_0^{+\infty}\frac 2{\mathrm{e}^x}\,\mathrm d x$ is convergent.