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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.