In a recent answer, I defined the function $\Psi(x)=\sum_{n=0}^\infty x^n\log(1+x^n)$ for $x\in(0,1)$ and proved the limit $\lim\limits_{x\to1^-}\left(\frac{1}{x}-1\right)\Psi(x)=2\log(2)-1$ by bounding it using the integral test as $$\color{blue}{-\dfrac{2\log(2)-1}{\log\left(x\right)}}\leq\color{purple}{\Psi(x)}\leq\color{red}{-\dfrac{\log\left(1+x^{-1}\right)\left(1+x\right)-1}{{x\log\left(x\right)}}}$$ However, I was asking myself (and now you): is there a way to express $\bf\Psi(x)$ using already known functions? I don't mind if the closed form uses standard special functions like Bessel functions, the incomplete Gamma function, elliptic integrals, $q$-analogs, etc. But I won't count as valid expressions that involve ad hoc functions (that is, special functions that aren't standard defined specifically to express this function in a closed manner). I know that its Taylor series is given by
$$\Psi(x)-\log(2)=\sum_{n=1}^\infty x^n\log(1+x^n)=\sum_{n=1}^\infty x^n\sum_{k=0}^\infty \alpha_kx^{kn}=\sum_{n=1}^\infty\sum_{k=0}^\infty\alpha_k x^{(k+1)n}=\sum_{n=1}^\infty\sum_{k=1}^\infty\alpha_{k-1}x^{kn}$$ $$=\sum_{m=1}^\infty\left(\sum_{d\vert m}\alpha_{d-1}\right)x^m=\sum_{m=1}^\infty\left(\sum_{1<d\vert m}\frac{(-1)^{d}}{d-1}\right)x^m\ \text{ where } \alpha_k:=\begin{cases}0 & \text{ if } k=0\\ \frac{(-1)^{k+1}}{k} & \text{ if } k\geq1\end{cases}$$ $$\boxed{(1)\ \Psi(x) = \log(2)+\sum_{m=1}^\infty\left(\sum_{1<d\vert m}\frac{(-1)^{d}}{d-1}\right)x^m\ }$$ but this leads to nothing since I don't know how to express the Dirichlet convolution involved in its terms. Also, as pointed out by @TymaGaidash, we can also express it as the following Lambert series $$\boxed{(2)\ \Psi(x) = \log(2)+\sum_{n=1}^\infty\frac{(-1)^n}{n(1-x^{-n-1})}}$$