A uniform bound if $f(x)\geq 1$ on $[0,\infty]$ Does there exists a constant $c>0$ such that whenever $f\in C[0,\infty]$ satisfies $f(0)=1$ and $f(x)\geq 1$ for all $x>0$ then
$$\int_0^\infty f^2(t)\ e^{-\int_0^t f(x)dx}dt <c.$$
Or can we construct such a sequence of $\{f_n\}$ such that the integral above increases to $\infty$?
Edit:  I think the sequence $f_n(x)=e^{x^n}$ is an example. 
$$\int_0^\infty e^{2t^n}e^{-\int_0^t e^{x^n}dx}dt=1+n\int_0^\infty t^{n-1}e^{t^n}e^{-\int_0^t e^{x^n}dx}dt$$
$$ =\dots=1+n!\int_0^\infty e^{t^n}e^{-\int_0^t e^{x^n}dx}dt=1+n!$$

Consider the series of constant functions $f_n(t) \equiv n$. Then, unless I'm mistaken,
$$
P[f_n] =\int_0^\infty \! n^2 \mathrm e^{-nt} \;\mathrm dt = \lim_{t\rightarrow \infty} n -(n\ \mathrm e^ {-nt }) = n,
$$
and thus $P$ is unbounded.
Edit: Note that I'm assuming $f^2(t)$ to mean $f(t)^2$, as in $\sin^2(t)$, and not $f(f(t)) = f\circ f(t)$.
Edit II (Correction): As pointed out below, I forgot about the condition $f(0)=1$. So to fix that, note that almost everything about $f_n(t)=n$ is alright except for the very first point, and an (arbitrarily small) neighborhood around it, that has little effect on the overall result. So what you need is function series $f_n(t)$ that satisfies $f_n(0)=1$, but that converges (over $n$) pointwise to $g_n(t) = n$ (the function that I mistakenly proposed above) on all $t > 0$. This means that for large $n$, $f_n$ is arbitrarily close to $g_n$, and almost only differs in the very first point.
An example of such a function would be
$$
 f_n(t) = (n-1)\operatorname{erf}(t\cdot n)+1,
$$
where $\operatorname{erf}(x)$ is the Gauss error function, here scaled to move from $1$ to $n$ over $t \in [1,\infty)$, at a rate that increases with $n$. For $n = \infty$, this $f_n(t)$ equals $g_n(t)$ everywhere but in the first point, which has no effect on the integrals.
