# “generalised” gamma-like integral $\int_0^\infty x^ne^{-f(n)x}dx$

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.