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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.

Tien Kha Pham
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    $f(t) = \frac{(1+t)^{\alpha}}{\ln^{\alpha}(1+t)}$, $\alpha>0$ ? – mastrok Apr 24 '17 at 09:35
  • $\frac{(1+t)^\alpha}{\ln^\alpha (t+1)}$ is actually greater than $(t+1)^\beta$ for some $\beta>0$ when $t$ is large enough. I'd like to see some "weaker" function. – Tien Kha Pham Apr 24 '17 at 16:55
  • For every $\beta>0$, $\frac{(1+t)^{\beta}}{\ln^{\beta}(1+t)}$ is weaker than $(1+t)^{\beta}$. – mastrok Apr 25 '17 at 05:52

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