# Possible limit involving the gamma function

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 $$n=100$$.
• 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). – Greg Martin Jun 30 at 0:01
• @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. – Iosif Pinelis Jun 30 at 0:45