Integral with gamma function

Question

Inspired by the limit

$$ \lim_{n\to \infty}\frac{\sqrt[n]{n!}}{n} = e^{-1} $$

and using desmos.com, I plotted the function

$$ f(x)=\frac{\sqrt[x]{\Gamma(x+1)}}{x}-\frac{1}{e} $$

and obtained the result bellow.

Don't know why, but I had the feeling that the integral

$$ \int_{0}^{\infty}\left(\frac{\sqrt[x]{\Gamma(x+1)}}{x}-\frac{1}{e}\right)dx $$

would converge. Then I used Wolfram to test it. Here's it's solution:

Using desmos.com, I started to test some values for the fraction $\frac{\sqrt[n]{n!}}{n}$ and for values greater than $n=169$, the computer can't solve, but I know that that the value will converge to $e^{-1}$, so large values of $n$ shouldn't be a problem. So the problem is the computer, which makes me think that Wolfram suffer the same type of issue.

So, is there an analytic way to prove that the integral converges? Does the integral have a closed form? If it does not converge, how can I prove it?

Thank you in advance.

Answer 1

As $x\to\infty$, the integrand behaves like (using Sterling’s formula) $$\frac{\left(\sqrt{2\pi x} (x/e)^x \right)^{1/x} }{x} -\frac 1e \\ = \frac{(2\pi x)^{\frac{1}{2x}} -1}{e} \\ = \frac{e^{\frac{\ln (2\pi x)}{2x}} -1}{e} \\ \approx \frac{\ln (2\pi x)}{2ex} $$

Evidently $\int_1^{\infty} \frac{\ln (2\pi x)}{2ex} dx$ diverges, so also the original integral.

Answer 2

There is also a problem around $x=0$ since $$\log (\Gamma (x+1))=\sum_{n=0}^\infty \frac{\psi ^{(n-1)}(1)}{n!} \,x^n=-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+O\left(x^4\right)$$ $$f(x)=\frac{\sqrt[x]{\Gamma(x+1)}}{x}-\frac{1}{e}$$ $$f(x)=\color{red}{\frac{e^{-\gamma }}{x}}+\left(\frac{e^{-\gamma }}{12} \pi ^2-\frac{1}{e}\right)+\frac{e^{-\gamma }}{2} \left(\frac{\pi ^4}{144}+\frac{\psi ^{(2)}(1)}{3}\right)x+O\left(x^2\right)$$ which is a quite good approximation.

Consider $$I_k=\int_{10^{-k}}^1 f(x)\,dx$$

$$\left( \begin{array}{ccc} k & \text{approximation}& \text{solution}\\ 1 & 1.35996 & 1.36433 \\ 2 & 2.66105 & 2.66542 \\ 3 & 3.95470 & 3.95907 \\ 4 & 5.24759 & 5.25197 \\ 5 & 6.54041 & 6.54478 \\ 6 & 7.83322 & 7.83759 \\ 7 & 9.12603 & 9.13040 \\ 8 & 10.4188 & 10.4232 \\ 9 & 11.7116 & 11.7160 \\ 10 & 13.0045 & 13.0088 \\ \end{array} \right)$$

Nice but diverging.