Does $f(x)\in L^1$ imply that $f'(x) \in L^1$? Let $f(x)$ be defined for all real numbers differentiable function of one variable.We know that:
$$\int_{-\infty }^{+\infty } |f(x)| \, dx\neq +\infty$$
Problem is to resolve if it is possible or not that:
$$\int_{-\infty }^{+\infty } |f'(x)| \, dx= +\infty$$
 A: Counterexample: take 
$$
f(x) = 
\begin{cases}
x^2e^{-x^2}\sin(1/x^2) & x \neq 0\\
0 & x = 0
\end{cases}
$$
Note that although $f$ is absolutely integrable,
$$
f'(x) = \begin{cases}
2e^{-x^2}\frac{x^2(x^2 - 1)\sin(1/x^2) + \cos(1/x^2)}{x}
& x \neq 0\\
0 & x = 0
\end{cases}
$$
Is not. 
In this way, I have used the function $e^{-x^2}$ to "play" with David's function and "redefine the tails".

Proof that $f'(x)$ is not absolutely integrable:
Note that $|f'(x)| \geq C \left|\frac{\cos(1/x^2)}{x}\right|$ for some constant $C$.   It is thus sufficient to show that the integral
$$
\int_{-1}^1 \left|\frac{\cos(1/x^2)}{x}\right| \,dx
$$
Diverges.  To show that this is the case, note that
$$
\int^{1/\sqrt{\pi n}}_{1/\sqrt{\pi(n+1)}} \left|\frac{\cos(1/x^2)}{x}\right| \,dx \\ \geq
\int^{1/\sqrt{\pi(n+1/4)}}_{1/\sqrt{\pi(n+3/4)}} \left|\frac{\cos(1/x^2)}{x}\right| \,dx \\ \geq
\left(\frac 1{\sqrt{(n + 1/4)\pi}} - \frac 1{\sqrt{(n + 3/4)\pi}}\right)\cdot \sqrt{n \pi} \cdot \frac{\sqrt{2}}{2}
$$
Show that this is bounded below by $D/n$ for some constant $D>0$.  From there, it follows that
$$
\int_{-1}^1 |f'(x)|\,dx \geq \sum_{n=-N}^N D/n
$$
From which we conclude that the integral diverges.
A: I think that trying to use a "single formula function" makes things unnecessarily complicated. Here is a counterexample. It is based on the function
$$
t\mapsto (x-a)^2(b-x)^2.
$$
This function, on the interval $[a,b]$ is differentiable with both the function and the derivative being zero at the endpoints. Such a feature allows us to put these "lumps" in different places and still get a differentiable function. Let
$$
f(t)=\sum_{n=1}^\infty n^8(x-n)^2(n+\frac1{n^2}-x)^2\,1_{[n,n+\frac1{n^2}]}.
$$
As $n$ grows, the lumps are thinner and higher. So we can get small areas with big derivatives:
$$
\int_{\mathbb R} |f|=\sum_{n=1}^\infty\int_n^{n+1/n^2}n^8(x-n)^2(n+\frac1{n^2}-x)^2
=\sum_{n=1}^\infty\,n^8\,\frac1{30n^{10}}=\sum_{n=1}^\infty\frac1{30n^2}<\infty.
$$
And
$$
\int_{\mathbb R}|f'|=\sum_{n=1}^\infty\int_n^{n+1/n^2}|2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)|\,dx\\
=\sum_{n=1}^\infty\int_n^{n+1/2n^2}2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)\,dx\\-\int_{n+1/2n^2}^{n+1/n^2}2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)\,dx\\
=\sum_{n=1}^\infty n^8\,\frac1{8n^8}=\infty
$$
A: Counterexample: $f(x)=\frac{\sin(e^x)}{1+x^2}$ - smooth function
$$\int_{-\infty}^{+\infty}\left|\frac{\sin(e^x)}{1+x^2}\right| \,dx\leqslant \int_{-\infty}^{+\infty}\frac{1}{1+x^2} \,dx=\pi$$
Which means that $f(x)\in L^1$
$$f'(x)=\frac{e^x \left(x^2+1\right) \cos \left(e^x\right)-2 x \sin \left(e^x\right)}{\left(x^2+1\right)^2}$$
Note: $\forall_{\varepsilon > 0} \lim_{X\to \infty } \, \int_X^{X+\varepsilon} f'(x) \, dx=0$
$$|f'(x)|=\frac{e^x \left(x^2+1\right) \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)^2}\geqslant \frac{1}{2}\frac{e^x \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)}\geqslant \frac{1}{4}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)}$$
$$\int_{-\infty}^{+\infty}|f'(x)| \,dx > \frac{1}{4}\int_{0}^{+\infty}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)} \,dx$$
$\left|\frac{2 x}{x^2+1} \sin \left(e^x\right)\right|\leqslant 1$ which means that if we prove divergence of $\int_{0}^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx$ we prove divergence of $\int_{-\infty}^{+\infty}|f'(x)| \,dx$.  $$\int_0^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx> \int_0^{+\infty}\cos \left(e^x\right)^2 \,dx=\int_0^{+\infty}\frac{1}{2} \,dx+\frac{1}{2}\int_0^{+\infty}\cos \left(2e^x\right) \,dx$$
Note that:$\left(\frac{e^x}{x^2+1}\right)^{'}=\frac{e^x (-1+x)^2}{\left(1+x^2\right)^2}\geqslant 0$, so finally if $\int_0^{+\infty}\cos \left(2e^x\right) \,dx$ is convergent than $\int_{-\infty}^{+\infty}|f'(x)| \,dx=+\infty$.
$$\int_0^{+\infty}\cos \left(2e^x\right) \,dx=\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$$
$\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$ is convergent by Dirichlet's test or even Leibniz test with $a_n=\int_{n\pi+0.5\pi}^{(n+1)\pi+0.5\pi}\left|\frac{\cos \left(v\right)}{v}\right| \,dv$ , which completes the proof.
