Is $\int_{0}^{\infty} \log (1+ 2\operatorname{sech}x)\,\mathrm dx$ convergent?

Question

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

Comparison test isn't helpful in finding the convergence, besides I can't really find the point of discontinuity.

Hint. For $x \geq 0$ we have $0 \leq \log(1 + 2\operatorname{sech} x) \leq 4 e^{-x}$. — Sangchul Lee, Sep 30 '17 at 06:29
As you see from the comments and answer the questions was solved, but maybe you want to see it with this code int log(1+2sech x)dx, from x=0 to infinite using Wolfram Alpha online calculator. Good luck. — , Sep 30 '17 at 07:57

score 2 · Accepted Answer · answered Sep 30 '17 at 07:27

For $x\to \infty$, Sangchul Lee gave the good hint.

Now, close to $x=0$, by Taylor $$\text{sech}(x)=1-\frac{x^2}{2}+O\left(x^4\right)$$ $$1+2\text{sech}(x)=3-x^2+O\left(x^4\right)$$ $$\log\left(1+2\text{sech}(x) \right)=\log (3)-\frac{x^2}{3}+O\left(x^4\right)$$

Is $\int_{0}^{\infty} \log (1+ 2\operatorname{sech}x)\,\mathrm dx$ convergent?

1 Answers1