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 $100.759456...$



Possible limit involving the gamma function
$\newcommand\Ga\Gamma$
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
$$1+\int\_0^1 g(x)\,dx=0.75330\ldots. $$




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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 $n=100$.


