We know that for all $\alpha > 0$, the functions $\dfrac{1}{(t+1)\cdot (t+1)^\alpha}$ belong to $L^1(0, \infty)$.
We also know that for all $\alpha > 0$, the functions $\dfrac{1}{(t+1)\ln^\alpha (t+1)}$ do not belong to $L^1(0, \infty)$.
My question is: is there any function $f(t)$ "in the middle of" $\ln^\alpha (t+1)$ and $(t+1)^\alpha$ such that $\dfrac{1}{(t+1)f(t)}$ belongs to $L^1$?
Let me state it mathematically:
Find a function $f: [0, \infty)\to (0, \infty)$ such that $\dfrac{1}{(t+1)f(t)} \in L^1(0, \infty)$ and $$\lim_{t\to \infty} \frac{f(t)}{\ln^\alpha (t+1)} = \lim_{t\to \infty} \frac{(t+1)^\alpha}{f(t)} = \infty, \quad \forall \alpha >0.$$
Thank you very much.