Does $ \lim_{n\to \infty}\int \dfrac{f_n(x)}{\pi^n}dx $ exist? Let $f_1(x)=e^x$ and define recursively 
$$
f_{n+1}(x)=(e^x)^{f_n(x)}.
$$
Does the following limit exist?
$$
\lim_{n\to \infty}\int \dfrac{f_n(x)}{\pi^n}dx
$$
 A: Assuming you are talking about limits of functions (the limit of antiderivatives to be exact), note that $(a^b)^c = a^{bc}$. Thus we are finding
$$ f(x) = \lim_{n \to \infty} f_n(x) =  \lim_{n \to \infty} \frac{1}{\pi^n} \int e^{xe^{nx}} dx $$
Fix a constant such that $f_n(x) = 0$ for each $n$, so that $f_n(x) = \frac{1}{\pi^n} \int_0^x e^{xe^{nx}} dx$. For $x > 1$, $e^{xe^{nx}} > e^{e^n}$. Thus on these values $f_n(x) \geq \frac{1}{\pi^n} (x-1) e^{e^n}$, which tends to infinity as $n$ tends to infinity. One can refine this method to show that the series diverges (converges to infinity) on $(0,\infty)$.
For $x < 0$, we note that $e^{xe^{nx}}$ is a decreasing sequence, hence the Lebesgue monotone convergence theorem implies that $\lim_{n \to \infty}\int_0^x \frac{e^{xe^{nx}}}{\pi^n}$ exists, and is equal to $\int_0^x \lim_{n \to \infty} \frac{e^{xe^{nx}}}{\pi^n}$. Now $e^{xe^{nx}}$ converges to zero for $x < 0$, for $nx \to -\infty$, hence $e^{nx} \to 0$, hence $\frac{e^{xe^{nx}}}{\pi^n} \to 0$. Thus for $x < 0$, $f(x) = 0$.
In conclusion, we find $f = \infty \cdot \chi_{(0,\infty)}$. To cheat a little, checking our answer is right, we look at the graph below for a large value of $n$:

