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.

• If it is of any relevance to you, the related function $\varsigma(s)=\prod_{n=2}^{\infty} \frac{1}{1-n^{-s}}$ has, in my opinion, some truly remarkable number-theoretical aspects. Namely that it can be written as the Dirichlet series $\sum_{n=1}^{\infty} \frac{\rho(n)}{n^s}$ where $\rho(n)$ is the so-called multiplicative partition function (see oeis.org/A001055). It is very simmilar to the usual partition function in that it counts the number of ways to factor a number. In fact, the product formula is very simmilar to the one for the generating function of the partition function. Aug 27, 2022 at 1:34
• @TianVlašić. Thanks. I've added the number theory tag. Aug 27, 2022 at 1:45
• Note that $f_m(x)$ itself seems to contain half the area of the rectangle $[0,1]×[0,1]$ as $m→∞$ and it is comparable with $\cos^2(πx/2 )$ Aug 27, 2022 at 2:05
• Each of the factors is integrable, continuous, and nonzero, so I think that ought to be true of the product as well, right? Certainly Desmos seems to believe it's integrable, not that Desmos is our most reliable guide. I don't get a lot of wiggling though, even out at $m=10^6$. Alas I do see a steady decrease of the integral as $m$ increases, which suggests the outlook for a lower bound may be hazy. Aug 27, 2022 at 3:34
• I've been just messing around with this... WolframAlpha says that for rational $x<\frac12$, the product converges, though the values it gives are... interesting. At $x=\frac12$ it converges to $\frac12$ exactly; at $x=\frac16$ it converges to... $(\cosh^2 (\pi\sqrt3/2))/6\pi^2$. I'd bet it converges for irrational values as well, though. Whether it converges for $x>\frac12$ is really unclear, though. Fractional powers don't like to behave much. Aug 27, 2022 at 7:50

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 $$0 < I < 1$$.

Can we get a decent approximation for I? - yeah, I think so.

The trapezoidal rule gives good results, because $$f^{(n)}(0^+)=f^{(n)}(1^-)=0$$ for each $$n>0$$.

Here are my computations, using the following PARI/GP script:

myexp(x)=if(x<-default(realbitprecision),0,exp(x));
foo(x)=myexp(suminf(n=1,(1-zeta(n/x))/n));
goo(n)=sum(k=1,n,foo((k-1/2)/n))/n;
experiment(n)={my(v=0.5);for(k=0,n,v=(v+goo(2^k))/2;print(v))};


\small\begin{aligned} \color{blue}{0.}\color{gray}{5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}\\ \color{blue}{0.49}\color{gray}{52524522182130053179347592830935038182739178227155813430014853398627049802084168378189484744909907}\\ \color{blue}{0.4960}\color{gray}{462685813435659657410735621739971142678008217517627767029731808108723102454366861311477695706614}\\ \color{blue}{0.49600}\color{gray}{13263546360125317620145429003484306469288432684171247707533969912066396804032830035759621991914}\\ \color{blue}{0.4960022}\color{gray}{198625044032046958121794546131346992908777037660564955370118897334400210136224769456663842563}\\ \color{blue}{0.49600222206}\color{gray}{45758305327177461401946937544240845348719200390983304017261198780787928938707341701795960}\\ \color{blue}{0.496002222066503}\color{gray}{9435348385778735042737942407442787721881965817897194427569958858545529012872132030474}\\ \color{blue}{0.4960022220665038758722}\color{gray}{291597400222685547259341485206377189559616296300739040313598923621170596850242}\\ \color{blue}{0.496002222066503875872217752736}\color{gray}{8950707201277324066195674090515553311094699224292721471410927775290176}\\ \color{blue}{0.49600222206650387587221775273692568333606393}\color{gray}{08945109817369766264054491087707420065200842499692225784}\\ \color{blue}{0.496002222066503875872217752736925683336063939363714480500547}\color{gray}{9605585454983581866471891249689081986411}\\ \color{blue}{0.496002222066503875872217752736925683336063939363714480500547711451723047013043337657919}\color{gray}{1646691496229}\\ \color{blue}{0.4960022220665038758722177527369256833360639393637144805005477114517230470130433376579194899311327705}\end{aligned}

• It's near from $$\int_{0}^{1}\operatorname{erf}\left(x\right)dx$$ . Aug 29, 2022 at 9:57
• Thanks. No direct hit on a lookup table. But that's not too surprising. Aug 29, 2022 at 15:51

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

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