$f(n)=$, for even integer’s $n$, $\lim_{n\to\infty} \int_{0}^{\infty} \frac{e^x}{^nx} dx$ $f(n)=$, for even integer’s $n$, $$\lim_{n\to\infty} \int_{0}^{\infty} \frac{e^x}{^nx} dx$$
$^n x$ is the tetrative function or ‘power tower’, which when $n=\infty$ can also be written in terms of y, as $x= e^\frac{ln(y)}{y}$, I mention this only because this may be doable in standard notation if taken in the dy world but I dont know how to effectively translate this equation.
For reference I am a sophomore in highschool, and havent been formally taught calculus, so my notation may not be perfect
Specifically the even $n$’s because any odd $n$ will not converge
I have a series of approximations which point towards an answer but I was curious if theres a was a way to either find an exact answer or if there was some special function or sum I was unaware of which could help
$f(2)\approx 6.34981$
$f(4)\approx 3.9184$
$f(6)\approx 3.74148$
$f(8)\approx 3.6939$
$f(10)\approx 3.6762$
$f(12)\approx 3.66844$
After this my wolfram alpha runs out of computation power and time, but it is clearly approaching some value.
Additionally the function can be wrote as just $$\lim_{n\to\infty} \int_{0}^{\infty} \frac{e^x}{^{2n}x} dx$$
for any integer $n$ but I left it as the original so you could better understand my framing when approaching the problem.
 A: We'll try to use the dominated convergence theorem to bring the limit into the integral and hopefully get something more manageable. For the boundedness part, it can be shown (like in this reference, for example) that $^{2n}x \ge 1/e$ for $x \in \left[0,e^{\frac1e}\right]$ and diverges to infinity for the other possible values.
Now, from the same reference we know that $^{2n}x$ converges to a branch of $y=x^{x^y}$ for $0<x<e^{-e}$. Using this post we can thus say that
$\frac{1}{^{2n}x}$ goes to $y=0$ for $x>e^{\frac1e}$, for $e^{-e}\le x \le e^{\frac1e}$ goes to $y=e^{W_0\left( -\ln(x)\right)}$ and for $0<x<e^{-e}$ it goes to $x = e^{y W_{-1}\left( \frac{-\ln(y)}{y}\right)}$, where $W_k(z)$ is the Lambert $W$ function.
This allows us to get an integral representation for your limit in terms of the $W$ function. By denoting $I_n = \int_{0}^{\infty}\frac{e^x}{^{2n}x}\mathrm{d}x$ we get
$$
I_n \to e^{1+e^{-e}}-1 - \int_{1}^{e} e^{ e^{y W_{-1}\left( \frac{-\ln(y)}{y}\right)}}\, \mathrm{d}y+\int_{e^{-e}}^{e^{\frac1e}} e^{x +W_0\left( -\ln(x)\right)} \,\mathrm{d}x \approx 3.67054
$$
where the last evaluation is obtained using W.A.

Alternatively, you can use the parametrizations also given in this question to get an integral representation for the value of the limit that doesn't use the $W$ function explicitly. Although this method does work, it results in very ugly integral expressions (specifically this one and this one), but this method also confirms the previous value of $3.67054...$ as the limit.
As a last bonus, it's easier to plot a parametric curve rather than a special function, so here's a graph of the previously given parametrizations alongside $\frac{e^x}{^{100}x}$:

