Is $\int_0^\infty \frac{dt}{e^t-xt}$ analytic continuation of $\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$?

The following power series apparently converges only for $-e \leq x <e$:

$$f(x)=\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$$

We can use it to define a real function $f(x)$, analytic in that interval.

However, we can also use an integral to define this function:

$$f(x)=\int_0^\infty \frac{dt}{e^t-xt}=\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$$

In the interval of convergence of the series these two definitions are equivalent. However, for $x<-e$ the power series diverges, but the integral converges:

enter image description here

Can the integral serve as the analytic continuation of $f(x)$ for $x<-e$? How to justify this?


The integral works great for the complex plane as well. Here I used the integral representation to plot the real and imaginary parts of $f(z)$:

enter image description here

We have problems on the real line for $x>e$. For the power series – they converge on a disk $|z|<e$ (with the boundary possibly included):

enter image description here

Solutions Collecting From Web of "Is $\int_0^\infty \frac{dt}{e^t-xt}$ analytic continuation of $\sum_{k=1}^\infty \frac{(k-1)!}{k^k} x^{k-1}$?"