<s>[INST] 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:
$$\int_{-\infty }^{+\infty } |f'(x)| \, dx= +\infty$$
 [/INST] 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.
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