If $f$ is Lebesgue integrable and $f''$ exists and is Lebesgue integrable, what can we say about the integrability of $f'$? Let $\newcommand{\RR}{\mathbb{R}}f: \RR \to \RR$ be a twice differentiable function that is in $\mathcal{L}^1$. Suppose that also $f'' \in \mathcal{L}^1$. Can we conclude that $f' \in \mathcal{L}^1$?
What about the case where we know $f \in \mathcal{L}^1$ and $f^{(m)} \in \mathcal{L}^1$, can we say something about the integrability of $f^{(k)}$ for $k \leq m$ ?
In general we can't say that the derivative of a smooth integrable function is integrable, but maybe the integrability of $f''$ helps?
 A: If $f''$ is continuous (or more generally, if $f'$ is absolutely continuous) then
$$
 f(x+1) = f(x) + f'(x) + \int_x^{x+1} (x+1-t)f''(t) \, dt
$$
from Taylor's theorem with the integral remainder. It follows that $f' \in {\cal L}^1$ if $$
 g(x) = \int_x^{x+1} (x+1-t)f''(t) \, dt
$$
is in ${\cal L}^1$ and that can be shown with the Fubini-Tonelli theorem:
$$
 \int_{-\infty}^{\infty} |g(x)| \, dx \le  \int_{-\infty}^{\infty}\left(\int_x^{x+1} (x+1-t)|f''(t)| \, dt \right) dx \\
=  \int_{-\infty}^{\infty} \left( \int_{t-1}^t (x+1-t)|f''(t)| \, dx \right) dt \\
= \int_{-\infty}^{\infty} |f''(t)|\left( \int_{t-1}^t (x+1-t) \, dx \right) dt \\
= \frac 12 \int_{-\infty}^{\infty} |f''(t)| \, dt < \infty \, .
$$
This answers the first part of your question.
The same approach works with higher derivatives: If $f$ and $f^{(n+1)}$ are in ${\cal L}^1$ and $f^{(n+1)}$ is continuous (or more generally, $f^{(n)}$ is absolutely continuous), then we can use Taylor's theorem for $n$ distinct non-zero offsets $0 < h_1 < h_2 < \ldots < h_n$:
$$
 f(x+h_j) = f(x) + \frac{f'(x)}{1!} h_j + \frac{f''(x)}{2!} h_j^2 + 
 \ldots + \frac{f^{(n)}(x)}{n!} h_j^n 
 + \int_x^{x + h_j} \frac{f^{(n+1)}(t)}{n!} (x+h_j - t)^n \, dt
$$
for $j=1, \ldots, n$. For a given $k$ with $1 \le k \le n$ one can find coefficients $c_1, \ldots, c_n$ such that
$$
 c_1 h_1^l + c_2 h_2^l + \ldots + c_n h_n^l = \begin{cases}
  1 & \text{ for } l=k \, , \\
  0 & \text{ for } l=1, \ldots, n,  l \ne k \, .
 \end{cases}
$$
That works because the determinant of the coefficient matrix of that linear equation system is $h_1 h_2 \cdots h_n$ multiplied with the Vandermonde determinant of $h_1, h_2, \ldots, h_n$ and therefore non-zero.
It follows that
$$
 \frac{f^{(k)}(x)}{k!} = \sum_{j=1}^n c_j \left( 
 f(x+h_j) - f(x) - \int_x^{x + h_j} \frac{f^{(n+1)}(t)}{n!} (x+h_j - t)^n \, dt
 \right)
$$
and the same reasoning as above shows that $f^{(k)} \in \cal L^1$.
