If $f \in C^\infty$, and $f$ is nonnegative and integrable in $\mathbb{R}$, can I say that $f^\prime$ is integrable? I'm not sure how to describe the question any further in the title than it is, but I will try to explain what I have done. If $f$ is a Schwartz function, I believe that $f^\prime$ will always be integrable, because the decay rate of $f$ is faster than any algebraic rate. Similarly, if $f$ was a $C_c^\infty(\mathbb{R})$ function, then clearly $f^{(k)}$ will be integrable because $f^{(k)}$ will be bounded and has compact support. So assume then that $f \in C^\infty (\mathbb{R})$, but is not a Schwartz function, nor has compact support. Can I still say that $f^\prime$ is integrable?
My intuition is that if $f$ is neither a Schwartz function of something with compact support, then $f$ must decay at some algebraic rate, $x^{-k}$ for $k>1$, since $f$ is also integrable itself. When I take the derivative of $x^{-k}$, I obtain an algebraic rate that decays even faster which seems to suggest that $f^\prime$ should also be integrable. Is this true? If so is there a rigorous proof of this? I cannot seem to find one, and if this is not true is there a counter example?
Edit: I forgot to add an extra constraint on $f$ in which I would also like $f$ to be strictly non-negative. My thoughts on this are I would like $f$ to be a density function or something like that.
 A: It's actually fairly easy to come up with a counter-example. We start with $f(x) = e^{-x^2}$. Now, as you know, this is a Schwartz function, so it doesn't do what we want, but we can "massage" it a bit and make its derivative large by adding very rapid oscillations to it, e.g. $g(x) = e^{-x^2}\sin(e^{x^2}x)$. 
$g'(x) = -2xe^{-x^2}\sin(e^{x^2}x) + (1+2x^2)\cos(e^{x^2}x)$, so $g'$ isn't integrable (assuming I computed the derivative correctly). But $g$ is as it's just the product of an $L^1$ function with an $L^\infty$ function.
EDIT (in response to OP's edit):
Here's what I'd try to do to make a non-negative function with your desired property.
Take $f_n$ to be a "spike" supported on $[n,n+1/n^3]$ of height $2n$. Then $\displaystyle\int_{\mathbb R} f_n = \frac12 \cdot \frac1{n^3} \cdot 2n = \frac1{n^2}$ such that $f = \sum_n f_n \in L^1(\mathbb R)$.
Next, convolve $f$ with a smooth mollifier. (And finally add $e^{-x^2}$ to make it strictly positive).
The resulting function will be smooth and $L^1$. See https://en.wikipedia.org/wiki/Mollifier & https://en.wikipedia.org/wiki/Convolution.
