Question: Consider $f(x)$ with $\displaystyle \int\_{-\infty}^{\infty}|f(x)|dx < \infty$. Does there exist some function , which doesn't has a bound and has this property? 

I thought no, but what about function which are near the zero and goes to infinite around zero?

Answer: 
There is. Take $$f(x)=\frac{1}{\sqrt[]{x}}\mathbf{1}\_{(0,1]}$$ This functions has no bound, but we have:
\begin{align}
\int\_\mathbb{R}|f(x)|\,dx=\int^1\_0 \frac{1}{\sqrt[]{x}}\,dx= 2<\infty
\end{align}
$$f(x)=\sum\_{n\geq 1} n e^{-n^6(x-n)^2}$$
is clearly positive, continuous and unbounded over $\mathbb{R}^+$ (since $f(n)\geq n$), but
$$ \int\_{0}^{+\infty} f(x)\,dx \leq \sqrt{\pi}\,\zeta(2).$$
Similarly, $\sum\_{n\in\mathbb{Z}}|n| e^{-n^6(x-n)^2}$ is a positive, continuous and unbounded function over $\mathbb{R}$ whose integral exactly equals $\frac{1}{3}\pi^{5/2}$.