If $f \in C^1(\mathbb{R}, \mathbb{R})$ and $|f(x)| + |x||f'(x)| \to 0$ as $|x| \to \infty$, do we have $f'(x) \in L^1(\mathbb{R})$? Let $f \in C^1(\mathbb{R}, \mathbb{R})$ such that
$$|f(x)| + |x||f'(x)| \to 0, \qquad \text{as $|x| \to \infty$}.$$

Is it necessary that $f'(x) \in L^1(\mathbb{R}; \mathbb{R})$ with respect to Lebesgue measure?

Setting $r = |x|$, it's clear that $|f'(x)|$ must decay faster than something like $1/((r + 1) \ln(r + 1))$ (for in that case $|f(x)| \gtrsim \ln(\ln(r +1)) $ as $r \to \infty$). On the other hand, $1/((r + 1) (\ln(r + 1))^{\rho})\in L^1[1,\infty)$ for any $\rho > 1$. So the question seems to be if there is a $C^1$ example, lying "in between" the examples just given, with $f' \notin L^1$.
Hints or solutions are greatly appreciated.
 A: A counter example is
$$f(x) = \int_x^{\infty} \frac{\sin t}{t \ln t} dt.$$
The integral converges by the alternating series test.
A: To add a few details to the accepted answer, for any $x > 0$, define the improper integral
$$
\int_{x}^\infty \frac{\sin t}{t\ln t}dt
$$
by
$$
g(x) = \int_{x}^\infty \frac{\sin t}{t\ln t}dt := \int_{x}^{\lceil x \rceil \pi } \frac{\sin t}{t\ln t}dt + \sum_{k = \lceil x \rceil}^\infty b_k,
$$
where $\lceil x \rceil$ denotes the ceiling function and
$$
b_k := \int_{k \pi}^{(k+1)\pi} \frac{\sin t}{t\ln t}dt.
$$
The series converges (by the alternating series test) because when $k$ is even
$$
0 \le \int_{k \pi}^{(k+1)\pi} \frac{\sin t}{t\ln t}dt \le \frac{\pi}{k\pi \ln(k \pi)},
$$
while if $k$ is odd
$$
0 \ge \int_{k \pi}^{(k+1)\pi} \frac{\sin t}{t\ln t}dt \ge \frac{-\pi}{k\pi \ln(k \pi)}.
$$
Note also that
$$
\left|\int_{x}^{\lceil x \rceil \pi } \frac{\sin t}{t\ln t}dt\right| \lesssim \frac{1}{\ln x}, \qquad \text{as $x \to \infty$.}
$$
When $x \notin \mathbb{N}$, $\lceil x \rceil$ is constant in a sufficiently small neighborhood of $x$, so it readily follows from the fundamental theorem of calculus that $g'(x) = -\sin x /(x \ln x)$. This continues to hold at $x \in \mathbb{N}$ by checking that the left and right hand derivatives agree (for instance, for the difference quotient $h^{-1}(g(x+ h) - g(x))$, $h > 0$, $g(x + h)$ has one fewer term in its $b_k$-series than $g(x)$, but by splitting up the integral coming from $g(x + h)$, all the constant terms cancel anyway). Thus $g \in C^1(0,\infty)$.
To finish, let $\chi \in C^\infty_0(\mathbb{R}; [0,1])$ with $\chi = 0$ near $[-1/2,1/2]$ and $\chi = 1$ on $(-\infty, -1] \cup [1, \infty)$. Now put
$$ 
f(x) = \begin{cases} 0 & -\tfrac{1}{2} \le x \le \tfrac{1}{2}, \\
-\chi(x) \int_{x}^\infty \frac{\sin t}{t\ln t}dt & x > \tfrac{1}{2}, \\ 
\chi(x) \int_{-\infty}^x \frac{\sin t}{t\ln t}dt & x < -\tfrac{1}{2}.
 \end{cases}
$$
where the improper integral in the last line is defined similarly to $g(x)$, and is shown to be in $C^1(-\infty, 0)$ along the lines as before. The support properties of $\chi$ ensure there are no convergence issues at zero. Thus $f \in C^1(\mathbb{R})$ with $f'(x) = \sin x/(x \ln x)$ when $|x|$ is large. Hence $f$ and $f'$ have the desired pointwise decay toward infinity, but $f' \notin L^1$.
