On $\int_0^1\frac{1}{x}\operatorname{frac}\left(\frac{1}{\operatorname{gd}(x)}\right)dx$, where $\operatorname{gd}(x)$ is the Gudermannian function

Question

Let $\operatorname{gd}(x)$ the Gudermannian function, see this MathWorld. And we denote with $ \left\{ x \right\} =\operatorname{frac}(x)$ the fractional part function that satisfies $$x=\left\{ x \right\} +\lfloor x\rfloor.$$

Question. How can you deduce if this integral $$\int_0^1\frac{1}{x}\left\{ \frac{1}{\operatorname{gd}(x)} \right\}dx$$ is convergent or divergent? Many thanks.

I know that $$\int_0^1\frac{1}{x}\frac{1}{\operatorname{gd}(x)}dx$$ does not converge, and that it is possible to write the integrand involving different functions, see previous reference.

Answer 1

We have $\text{gd}(x) \sim x$ as $x\to 0$ so the integral is divergent by a comparison with the integral $$\int_{0}^1\frac{1}{x}\left\{\frac{1}{x}\right\}{\rm d}x \underbrace{=}_{\text{substitution } u = \frac{1}{x}} \int_1^\infty \frac{\{u\}}{u}{\rm d}u = \sum_{n=1}^\infty \int_n^{n+1} \frac{x-n}{x}{\rm d}x = \sum_{n=1}^\infty 1 - n\log\left(1 + \frac{1}{n}\right)$$ which diverges by comparison to the harmonic series as $1 - n\log\left(1 + \frac{1}{n}\right) \sim \frac{1}{2n}$. for $n\to\infty$