What can we say about $f(x)=\prod_{n=2}^\infty(1-n^{-1/x})$? Main Question:
What can we say about
$f(x)=\prod_{n=2}^\infty(1-n^{-1/x})$?
Is $f(x)$ integrable from $0$ to $1$? Is it continuous? If we have an affirmative answer to the question on integrability...
$$I=\int_{0}^{1}f(x)\,dx=\int_{1}^{\infty}\frac{1}{x^2}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{x}}\right)dx$$
Can we get bounds on $I$? Can we get a closed form for I? Can we get a decent approximation for $I$?
Motivations
I saw this post wherein I found
$$ \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^p}\right) = \prod_{\omega : \omega^p = 1} \frac{1}{\Gamma(2-\omega)}. $$
After plotting $f(x)$, I found myself unsatisfied when I couldn't get a handle on $I$ using desmos. Maybe the issue is that $f_m(x)=\prod_{n=2}^{m}(1-n^{-1/x})$ don't converge fast enough? As I run $m\to \infty$ I observe $I_m= \int_0^1 f_m(x)\,dx$ wiggling.  I'm not quite sure how to proceed in my curiosities.
Thanks for any insights.
 A: Let $x>1$.  Then $\sum_{n=2}^\infty 1/n^x$ converges (to $\zeta(x)-1$). The terms $1/n^x$ are nonnegative. Therefore, the infinite product $\prod_{n=2}^\infty(1-1/n^x)$ converges absolutely.  (And, in particular, it does not diverge to zero.)  We have
$$
0 < \prod_{n=2}^\infty\left(1-\frac{1}{n^x}\right)< 1,
$$
Thus the integral
$$
I=\int_1^\infty \frac{1}{x^2}\prod_{n=2}^\infty\left(1-\frac{1}{n^x}\right)\;dx
$$
converges by comparison with $\int_1^\infty\frac{1}{x^2}\,dx$.
The estimate we get from this is $0 < I < 1$.
A: Using the work @metamorphy we have :
$$\operatorname{erf}\left(1\right)+\frac{\left(e^{-1}-1\right)}{\sqrt{\pi}}-\frac{\left(e^{-1}-1\right)^{20}}{\sqrt{\pi}}-\frac{\left(1-e^{-1}\right)^{25}}{\sqrt{\pi}}+1/100<I<\int_{0}^{1}\operatorname{erf}\left(x\right)dx+1/100$$
Perhaps someone can show it or refine it to give an infinite series .
Hope it helps ;-)
Edit same day :
We can use a life to finding it .I let you one example :
$$\frac{2\pi\left(\gamma-10\right)}{83+63\gamma\cdot}$$
Where there is the Euler-Mascheroni constant .
Found semi-manually (WA+Desmos)
Edit 30/08/2022 :
It's the more concise approximation I can find :
$$\sqrt{\frac{2011\cdot C_{len}}{2840\cdot\sqrt{10}}}$$
Where there is the Lengyels constant https://mathworld.wolfram.com/LengyelsConstant.html
