What is the value of $\int_0^{\infty}\frac{1}{x!}\,dx$

Question

I'm trying to figure out $\int_0^{\infty}\frac{1}{x!}\,dx$, but with no success, I tried approximating using Simpson's rule but i got $\frac{4e^2-e+2}{6e}$ which is about 1.76814... but wolframalpha says 2.26653, so what is the value of that integral?

Answer 1

For $x \geq 1$ the simplest upper bound is $e-1$, and the lower bound is $e-2$, because of the convergent sum $\sum_{k=1}^{\infty}\frac{1}{k!}$

Answer 2

Unless I'm mistaken, there's no neat solution to the integral (and this is assuming exchanging $x!$ with $\Gamma(x+1)$).

The constant, $2.2665345076\dots$, however, has some very neat properties, as described on the OEIS. For instance, Another way to represent such constant, Discovered by Ramanujan, is

$$e-\int_{-\infty}^\infty e^{\frac{-e^x}{x^2+\pi^2}}dx$$

It's also described as

a variation of the Fransén-Robinson constant. $$F=\int_0^\infty\frac{1}{\Gamma(x)}dx\approx2.80777$$

The relationship is clearly trivial.

Edit: Also, when it comes to estimation and evaluation, apparently the difference of these two constants "[...] may be accurately estimated by using bounds given in comments of A268896"

Answer 3

If you want to use the factorial and Simpson rule (then a step of $1$) you have $$\int_0^{\infty}\frac{1}{x!}\,dx=\frac 16 \sum_{a=0}^\infty \left(\frac{1}{(a+1)!}+\frac{4}{\left(a+\frac{1}{2} \right)!}+\frac{1}{a!} \right)$$ and you see that you have to work with non-integer values of the factorial function (then, the need of the gamma function as said in comments and answers).

This would give, as an approximation, $$\int_0^{\infty}\frac{1}{x!}\,dx\sim \frac 16 \left(4 e \,\text{erf}(1)+2 e-1\right) \approx 2.26656$$ which is not bad at all.

If you do not want to use non-integer factorials, you can bound the summation by $$\frac{5}{(a+1)!}+\frac{1}{a!}<\frac{1}{(a+1)!}+\frac{4}{\left(a+\frac{1}{2} \right)!}+\frac{1}{a!}< \frac{1}{(a+1)!}+\frac{5}{a!}$$ which makes $$e-\frac 56 <\int_0^{\infty}\frac{1}{x!}\,dx <e-\frac 16$$ and the average value of the bounds is $e-\frac 12 \approx 2.21828$