TITLE: Anything about $\prod_{n \ge 1} (1 + n^{-n})$?
QUESTION [6 upvotes]: Sophomore's dream is especially the statement that the sum, let me call it $s$, of the (convergent) real series $\sum_{n \ge 1} n^{-n}$ is equal to the (improper) integral $\int_0^1 x^{-x} dx$. A few digits of the decimal representation of $s$ are recorded on the OEIS. A couple of days ago, a user of an Italian math forum asked for information about a multiplicative analogue of $s$, namely the constant $p := \prod_{n \ge 1} (1 + n^{-n}) \approx 2.60361$. After browsing through Finch's Mathematical Constants (Cambridge University Press, 2003), I couldn't find anything there about $p$, but on the other hand, I couldn't find anything about $s$ either. So I guess my question is: 

Do you have any pointer to a piece of literature where $p$ pops up in connection to some (interesting) math?

REPLY [3 votes]: Logarithm $\log p$ of this product is also some definite integral (not a surprise, any number is a definite integral of appropriate function), but the integrand is more sophisticated than $x^{-x}$. 
Logarithm of your product equals 
$$
\log p=\sum \log(1+n^{-n})=\sum_n \sum_k(-1)^{k-1} \frac{n^{-kn}}{k}=\sum_k \frac{(-1)^{k-1}}{k} c_k, 
$$
where $$c_k:=\sum_{n\geq 1} n^{-kn}.$$
We have 
$$
\int_0^1 x^a(-\log x)^bdx=\frac{b!}{(a+1)^{b+1}}.
$$
hence $$c_k=\int_0^1 \sum_n \frac1{(kn-1)!}x^n(-\log x)^{kn} dx=\int_0^1 x^{1/k}g_k(-x\log x) dx,$$
where $$g_k(t)=\sum \frac{t^{kn-1}}{(kn-1)!}=k^{-1}\sum_{w:w^k=1} we^{wt}.$$
Thus $$c_k=\frac1k\int_0^1x^{1/k} \sum_{w:w^k=1} w x^{-wx}.$$
Therefore $\log p=\int_0^1 g(x)dx$, where
$$
g(x)=\sum_{k,w:w^k=1} \frac{(-1)^{k-1}}{k^2}x^{1/k}wx^{-wx}. 
$$
We may change order of summation, fixing $w=e^{2\pi ai/b}$ for coprime $a,b$. Then it arises for $k=b,2b,\dots$. We see that coefficient of $wx^{-wx}$ equals
$$
b^{-2}\sum_{k=1}^{\infty}(-1)^{kb-1} \frac{x^{\frac1{bk}}}{k^2}.
$$
I do not recognize this function, but maybe it may in turn be expressed via some generalized polylogarithms or whatever.