I would like to prove that the following integral is convergent
$$ \int_{0}^{+\infty} \frac{T(f)(t)}{(1+t)^2} \; dt$$
with $\forall t \geq 0, \; \displaystyle T(f)(t) = \int_{0}^{t} \frac{\ln(1+x)}{x} \, dx.$
Using integration by parts, I was able to prove that, for $M>0$ :
$$ \int_{0}^{M} \frac{ T(f)(t) }{ (1+t)^2 } \; dt = - \frac{T(f)(M)}{1+M} + \int_{0}^{M} \frac{ \ln(1+t) }{ t(1+t) } \; dt. $$
The idea would be to let $M \to +\infty$.
Given that $t \mapsto \displaystyle \frac{ \ln(1+t) }{ t(1+t) }$ is integrable on $\mathbb{R}^{+}$, I only have to show that $\displaystyle \frac{ T(f)(t) }{1+t}$ has a limit as $t \to +\infty$. However, I was not able to prove this. I was only able to prove that :
$$ \forall t > 0, \; \frac{1+t}{t} \ln(1+t) - 1 \leq T(f)(t) \leq t $$
but this is not enough to show that $\displaystyle \lim \limits_{t \to +\infty} \frac{ T(f)(t) }{1+t}$ exists.
Hint: $\int_0^1 \frac{\ln(1+x)}{x} \, dx \le 1.$ So for $t \ge 1,$
$$\int_0^t \frac{\ln(1+x)}{x} \, dx \le 1 + \int_1^t \frac{\ln(1+x)}{x} \, dx \le \ln(1+t)\int_1^t \frac{1}{x} \, dx = \ln(1+t)\cdot \ln t.$$