# Error Function limit

$$\prod_{n=1}^{\infty}{\frac{2}{\sqrt{\pi}}\int_0^n e^{-x^{2}} \mathrm{d}x} \approx 0.83874$$

Is it a known constant? I couldn't find anything about it. Do you know ways to calculate the value efficiently?

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An interesting(?) but unfortunately totally irrelevant observation: Googling 0.83874 yields about 12600 hits. – Harald Hanche-Olsen Nov 6 '12 at 16:04

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.
Could you tell me, how you did this? $$\int_n^\infty e^{-x^2}\,dx=\frac{e^{-n^2}}{2n}+\frac12\int_n^\infty\frac{e^{-x^2}}{x^2}\,dx$$ – Gunnar Nov 6 '12 at 16:29
@Gunnar: Write the integrand as $xe^{-x^2}\cdot x^{-1}$ and do a partial integration, integrating $xe^{-x^2}$ to get $-\frac12e^{-x^2}$ and differentiatiing $x^{-1}$ to get $-x^{-2}$. – Harald Hanche-Olsen Nov 6 '12 at 16:54