Sum of $\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}$

Question

I want to find the sum of the following series

$$\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}$$

Using theorems on integration and differentiation of series. I can set $t=\ln x+1$ so that I get $$\sum_{n \geq 1} \frac{t^n}{n^n}$$

But then I don't see how to proceed, since that $n^n$ is difficult to see as the result of a differentiation or integration. How can I see it?

Answer 1

Only a hint:

For $\,a\geq 0\,$ it's $$\sum\limits_{k=0}^\infty\frac{x^k}{(ak+1)^{k+1}}=\int\limits_0^1 t^{-xt^a}dt $$

and the base for this result is calculating $\enspace\int\limits_0^1 t^b(\ln t)^cdt\enspace$.

Therefore with $a:=1$ and $x$ substituted by $\ln x+1$ we get $$\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}=(\ln x+1)\int\limits_0^1 t^{-t(\ln x +1)}dt \enspace.$$