Existence of a class of function 
Consider a function $\gamma: [0, \infty) \to (0, \infty)$ satisfying these following conditions:
i) $\gamma$ decreases on $[0, \infty)$ and $\displaystyle \lim_{t\to \infty} \gamma(t) = 0$;
ii) $\gamma\not \in L^1$;
iii) $t\dot \gamma \in L^1$.

By part integration, we get $$\int_0^T t\dot \gamma(t) dt = T\gamma(T) - \int_0^T \gamma(t)dt.$$
Since $t\dot \gamma \in L^1$ and $\gamma \not \in L^1$, we conclude $\displaystyle \lim_{t\to \infty} t\gamma(t) = \infty$.
I could not find any $\gamma$, so I wondered if such a function exists.
Thank you very much for any hint or solution.
 A: Such functions do not exist.
If $t \mapsto t\dot{\gamma}(t)$ belongs to $L^1$, then $\dot{\gamma}$ is locally integrable. Since $\lvert\dot{\gamma}(t)\rvert \leqslant \lvert t\dot{\gamma}(t)\rvert$ for $t \geqslant 1$, $\dot{\gamma}$ is integrable over $[1,+\infty)$, and since $\dot{\gamma}(t) \leqslant 0$ and $\gamma(0)$ is finite, $\dot{\gamma}$ is also integrable over $[0,1]$, so it follows that $\dot{\gamma} \in L^1([0,+\infty))$. Now we have
$$\gamma(x) = -\int_x^{+\infty} \dot{\gamma}(t)\,dt = \int_x^{+\infty} \lvert \dot{\gamma}(t)\rvert\,dt,$$
and hence
\begin{align}
\int_0^{+\infty} \gamma(x)\,dx &= \int_0^{+\infty} \int_x^{+\infty} \lvert\dot{\gamma}(t)\rvert\,dt\,dx \\
&= \int_0^{+\infty} \int_0^t\,dx\, \lvert\dot{\gamma}(t)\rvert\,dt \\
&= \int_0^{+\infty} t\lvert\dot{\gamma}(t)\rvert\,dt \\
&< +\infty
\end{align}
by Tonelli's theorem.
Thus $i)$ and $iii)$ imply $\gamma \in L^1([0,+\infty))$.
A: As a complement to Daniel Fischer's answer, note that if you restrict your domain to $(0,\infty)$ (or increase your codomain to $(0,\infty]$) then such functions do exist, one example being
$$\gamma(t)=\begin{cases}
t^{-1/2}&\text{if }0<t<1,\\
t^{-2}&\text{if }t\ge1.
\end{cases}$$
You can easily construct a smooth version if you prefer, since all that matters is what happens at $0$ and $+\infty$.
