Integral inequality similar to Hardy's inequality I am not very sure if that's research level, I hope you don't find it too elementary for this place.
I am trying to solve the following puzzle:
We are given a real function $f$, where $f(x) \geq 0$ and $F := \int_0^x f(t) dt$ and some real $p>1$.
Does $\int_0^\infty f(x)^p e^{-x}dx < \infty$ imply $\int_0^\infty F(x)^pe^{-x}dx < \infty$ ? 
My current approach was modifying proofs of Hardy's theorem. This approach has not been very successful. I keep running into hard to compute integrals that I would need to bound. I hope somebody could maybe provide me with some direction in which I should look. If somebody was interested I could provide details of calculations so far.
 A: Here is an idea and I will leave details to you (it might be that I made a stupid mistake somewhere therefore these computations must be checked carefully). All functions below should be sufficiently nice so that all formulas make sense. First I will formulate a lemma. 
Lemma:
Let $p>1$. If $\psi(0)=0, \psi' >0$ and $\psi(\infty)=\infty$ then for any $f\geq 0$ and any (nice) measures $d\mu(t)$ we have 
$$
\int_{0}^{\infty}\left(\int_{0}^{t}f(s)ds\right)^{p}d\mu(t)\leq \int_{0}^{\infty}f(y)^{p} \left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]dy \quad (1)
$$
Remarks:
1) We have a freedom of choosing $\psi$. One can optimize the quantity $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]$ in the right  hand side of (1) over all $\psi$ and find the best bounds (This becomes some ODE problem which should not be a difficult problem so that I will leave details to the readers).
However, for our purposes we do not need to know the best $\psi$. One just needs to guess the right one to find some bound. Let me show you how it works on your example. In your case $d\mu(t)=e^{-t}dt$. Instead of $\psi$ take something like $\psi(t)=te^{\frac{t}{10(p-1)}}$. Then $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right] \leq C e^{-y}$. So the claim follows. 
Proof of Lemma:
Let $\psi$ be a convex function, and let $w$ be  some positive function (we will choose them later). Then Jensen's inequality together with Fubini's theorem implies:
\begin{align*}
&\int_{0}^{\infty} \varphi\left( \frac{1}{x}\int_{0}^{x} g(s)ds\right) w(x)dx\leq \int_{0}^{\infty}\frac{w(x)}{x}\int_{0}^{x}\varphi(g(s))dsdx=\\
&\int_{0}^{\infty} \varphi(g(s))\int_{s}^{\infty}\frac{w(x)}{x} dxds
\end{align*}
Let $\varphi=u^{p}$ and lets make a change of variables $s=\psi(y)$ and $x=\psi(t)$. Then the above inequality takes the form:
\begin{align*}
\int_{0}^{\infty} \left(\int_{0}^{t} g(\psi(y))\psi'(y) ds\right)^{p} \frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt\leq
\int_{0}^{\infty} [g(\psi(y))\psi'(y)]^{p}\int_{y}^{\infty}\frac{w(\psi(t)) \psi'(t)}{\psi(t) (\psi'(y))^{p-1}} dtdy
\end{align*}
Now choose $g$ so that $g(\psi(y))\psi'(y)=f(y)$. And choose $\frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt = d\mu(t)$ (for example take $w(\psi(t))dt =\frac{\psi^{p}(t)d\mu(t)}{\psi'(t)}$ ). Then the lemma follows. 
A: The question has been already solved by Paata Ivanisvili. I just want to share my insight.
I think I have found the solution. I will use generalized Hardy's inequality, which unfortunately is a bit like killing a fly with a cannon.
As a source I will cite http://www.encyclopediaofmath.org/index.php/Hardy_inequality where you can find references to the original source. 
Lemma - generalized Hardy's (cited from Encyclopedia of Math).
$$
\int_0^\infty \left| \phi(x) \int_0^x f(t) dt \right|^p dx \leq C \int_0^\infty \left| \psi(x)  f(x) \right|^p dx
$$
$\iff$
$$
\sup_{x > 0} \left[ \int_x^\infty  | \phi(t) |^p dt \right]^{\frac{1}{p}} \left[ \int_0^x | \psi(t) |^{-q} dt \right]^{\frac{1}{q}} < + \infty,
$$ where $\frac{1}{p} + \frac{1}{q} = 1 $. 
After using this inequality, the task comes down to verification of the assumption, which is pretty easy. Let me verify.
Given a function $f \geq 0$ and $F(x) = \int_0^{x} f(t)
dt$ I claim  $\int_0^\infty f(x)^p e^{-x}dx < \infty$ implies
$\int_0^\infty F(x)^pe^{-x}dx < \infty$. 
Proof: Assume $\psi(t) = \phi(t) = e^{- \frac{t}{p} }$. When
$\frac{1}{p} + \frac{1}{q} = 1 $, then $\frac{1}{q} = \frac{p-1}{p}$ and $-q = \frac{p}{1-p}$. We investigate
$$
\sup_{x > 0} \left[ \int_x^\infty  e^{-t} dt \right]^{\frac{1}{p}} \left[
\int_0^x e^{-t \frac{1}{1-p}} dt \right]^{\frac{p-1}{p}} = \sup_{x > 0} e^{- \frac{x}{p}} \left( e^{\frac{x}{p-1}} - 1 \right)^{\frac{p-1}{p}}
=
$$
$$
\sup_{x > 0} \left( e^{- \frac{x}{p-1}} e^{\frac{x}{p-1}} - e^{- \frac{x}{p-1}} \right)^{\frac{p-1}{p}}
=
\sup_{x > 0}
    \left(
        1 -
    e^{- \frac{x}{p-1}}
    \right)^{\frac{p-1}{p}},
$$
which is bounded as $p>1$. Now we apply Hardy and our theorem is verified.
