Assigning a value to the divergent sum of the factorial

Question

As seen here, it is possible to assign a value to the alternating factorial series:

$$\sum_{n=0}^\infty(-1)^nn!\approx0.596$$

I am interested in the non-alternating series. Here is my attempt:

$$\sum_{n=0}^\infty n!=\sum_{n=0}^\infty\int_0^\infty x^ne^{-x}dx$$

$$=\int_0^\infty\sum_{n=0}^\infty x^ne^{-x}dx$$

$$=\int_0^\infty\frac{e^{-x}}{1-x}dx$$

However, there is a singularity at $x=1$, so I cannot proceed any further with this method.

I am wondering if someone could finish the last line or provide insight into a possible other method to solving this problem.

Answer 1

You could interpret this as principal value: $$ \lim_{\epsilon\rightarrow 0^+} (\int_0^{1-\epsilon} + \int_{1+\epsilon}^\infty) \frac{e^{-t}}{1-t} dt$$ So the integral evaluates to something like 0.6971796...

Answer 2

This question could use a little more attention. I believe the standard value for $\sum_{n=0}^{\infty} n! = -e^{-1}\text{Ei}_1\left( {-1} \right)$ see here where it is shown that

$$ \sum_{n=0}^{\infty} n!x^n = -\frac{1}{x} e^{-\frac{1}{x}} \text{Ei}_1 \left( {-\frac{1}{x}} \right) $$

Setting $x=1$ gives you the desired result. Numerically this appears to be about 0.0807 as per here