I have noticed that if we have an integral of the form: $$I[f]=\int_0^\infty x^ne^{-f(n)x}dx=\frac{1}{f^{n+1}(n)}\int_0^\infty x^ne^{-x}dx=\frac{n!}{f^{n+1}(n)}$$ I was wondering what kind of restrictions would need to be applied to $f$ in order for this integral to converge for all values of $n$. An obvious one to me is that $f(n)>0$ and for it to converge for $n\to\infty$ we would require $$f(n)>(n!)^{\frac 1{n+1}}\tag{1}$$ are there any other requirements or functions people can think of that fit these requirements? Thanks
some obvious ones to me are: $f(n)=n!,(an)!$ but could an exponential work?
In terms of satisfying $(1)$ using sterlings approximation as suggested we get $$(n!)^{\frac{1}{n+1}}\approx (2\pi)^{1/2(n+1)}\frac{e^{1/(n+1)}n}{n^{1/2(n+1)}}$$ and as this approaches infinity I believe it is approx to $n$ so I think: $$f(n)=|n|^\alpha+a$$ would converge for all values $n$ where $\alpha>1,a>0$
The integral converges as long as $n > -1$ and $f(n) > 0$. It diverges at $0$ if $n \le -1$, and diverges at $\infty$ if $f(n) \le 0$ and $n \ge -1$.
If you want $\lim_{n \to \infty} \frac{n!}{f(n)^{n+1}} = 0$, noting that $n! \sim \sqrt{2\pi} n^{n+1/2} e^{-n}$ by Stirling's approximation, $f(n) = n/e$ would work, while $f(n) = t n $ for $0 < t < 1/e$ would not.