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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 $100.759456...$

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$\newcommand\Ga\Gamma$ 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 $n=100$.

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    $\begingroup$ 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
    Commented Jun 30, 2020 at 0:01
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    $\begingroup$ @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
    Commented Jun 30, 2020 at 0:45

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