Borel sum of $ 1!+2!+3!+.... $

Question

I know that the Borel sum of $ \sum_{n=0}^{\infty}(-1)^{n}n! $ is $ \int_{0}^{\infty} dx \frac{e^{-x}}{1+x} $

but what happens with the sum $ \sum_{n=0}^{\infty}n! $

the Borel sum should be $ \int_{0}^{\infty} dx \frac{e^{-x}}{1-x} $ which has a pole at $ x=1 $ using Shothotsky's formula I get

$$ PV \int_{0}^{\infty} dx \frac{e^{-x}}{1-x}-i\pi e^{-1} $$

however this is a complex number.

Answer 1

We have asymptotic series $$ e^{-x} \text{Ei}(x) \sim \sum_{j=0}^\infty \frac{j!}{x^{j+1}} \qquad \text{as } x \to +\infty $$

We can interpret "Borel summation" as: "Let's plug in $x=1$, even though the series diverges there!" $$ e^{-1}\text{Ei}(1) \approx 0.69717488 $$

Answer 2

(This should be a comment to @GEdgar's answer, but does not fit the input-box)

In an article we find a curious discussion about Maple summing this series to a complex value where the real value is that which GEdgar points out.

I just insert the screenshot (don't have the software to extract the text nicely):
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It reminds the reader, that things are far from trivial...