$\prod_{n=1}^{\infty}{\frac{2}{\sqrt{\pi}}\int_0^n e^{-x^{2}} \mathrm{d}x} \approx 0.83874 $
Is it a known constant? I couldn't find anything about it. Do you know ways to calculate the value efficiently?
$\prod_{n=1}^{\infty}{\frac{2}{\sqrt{\pi}}\int_0^n e^{-x^{2}} \mathrm{d}x} \approx 0.83874 $
Is it a known constant? I couldn't find anything about it. Do you know ways to calculate the value efficiently?
I don't know the constant, but when it comes to calculating the product, we might note that it can be written $\prod_{n=1}^\infty\Bigl(1-\frac{2}{\sqrt{\pi}}\int_n^\infty e^{-x^2}\,dx\Bigr).$ When $n$ grows large, partial integration yields $\int_n^\infty e^{-x^2}\,dx=\frac{e^{-n^2}}{2n}+\frac12\int_n^\infty\frac{e^{-x^2}}{x^2}\,dx,$ which gives you a reasonable handle on the integral on the left. Taking the log of the infinite product, you get a sum in which this will give you a somewhat decent approximation for the tails of the sum.
There are many details to fill in, but off the top of my head this outlines how I would go about computing the product.