Does $$\lim_{n \to \infty} \int_{0}^{1} \Gamma(x)^{n/(n+1)}dx - n$$ exist?

Here's some background. The integral

$$\int_{0}^{1} \Gamma(x) dx$$

diverges rather slowly. Inserting the exponent $n/(n+1)$ perhaps leads to a nice surprise---that the floor of resulting integral appears to be $n$. For example, for $n = 100$, the integral has a value of $100.759456...$


$\newcommand\Ga\Gamma$ Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and $-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$.

So, the limit in question is $$1+\lim_n J_n,$$ where $$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$ $$f_n(x):=g(x)-h_n(x),$$ $$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$ Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is $$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$

As seen from the proof, the rate of convergence here is $O(1/n)$. So, the limit value $0.75330\ldots$ is in agreement with the value of the integral you computed for $n=100$.

| cite | improve this answer | |
  • 2
    $\begingroup$ More generally, if $h(x)$ is any function on $(0,1]$ such that $h(x)-\frac1x$ is continuous at $x=0$, then $\int_0^1 h(x)^{n/(n+1)}\,dx - n \to \int_0^1 \big( h(x)-\frac1x \big)\,dx$. The proof of this more general statement along the lines of your solution is even cleaner (it shows that the functional equation for $\Gamma$ is irrelevant, for example). $\endgroup$ – Greg Martin Jun 30 at 0:01
  • 2
    $\begingroup$ @GregMartin : The above proof will indeed hold if you write $h(x)$ and $xh(x)$ instead of $\Gamma(x)$ and $\Gamma(1+x)$, respectively, everywhere in the proof, assuming that $h(x)-1/x$ is bounded. $\endgroup$ – Iosif Pinelis Jun 30 at 0:45

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.