give an example of a function that is integrable in $\mathbb R $ and $\lim_{ x\to \infty}f(x)\neq0$ i did a search for such function but didn't found anything useful/complete ! , like this : 
Integrable function $f$ on $\mathbb R$ does not imply that limit $f(x)$ is zero
is there any function that is integrable and $\lim_{x \to \infty}f(x) \neq0 $ and $\infty$ ??
 A: If the limit $L:=\lim_{x\to\infty} f(x)$ exists and is nonzero, then surely
$\int_0^b f(x)\,dx$ grows essentially like $Lb$ as $b\to\infty$ (because for big $b$, $\int_{b}^{b+1} f(x)\,dx\approx \int_b^{b+1}L\,dx$). Note that the question you linked to talks about the $\limsup$, not the $\lim$.
A: Take a function whose graph is a sequence of triangles whose bases are the $x$ axis, and the $n$-th triangle has size of $\frac1{n^2}$.
The integral of this function is finite, but there is no limit at $\infty$.
A: YOu can do this with a picture.  At each integer, draw a bump with area $1/2^n$ under it.  This gives the graph of a function $f$ with 
$$\int_0^\infty f(x)\,dx = 1$$
but you do not have $f(x) \to 0$ as $x\to\infty$.  In fact, you can draw these bumps as tall as you would like so you could have
$$\limsup_{x\to\infty} f(x) = +\infty.$$
A: $f(x) = \sum_{n=1}^\infty 1_{[n,n+n^{-2}]}(x)$
A: You can also form a function using the geometric series.  So let $f_n = \chi_{[n,n+ \frac{1}{2^n}]}$ and set $f = \sum_{n=1}^{\infty} f_n$.  $f$ is 1 infinitely often so it doesn't tend to 0, but it's clearly integrable.
A: If we're talking about Riemann integration, then we can also take $$f(x)=\sin(x^2).$$
