Here is a possible explicit series expansion for an inverse of the Exponential Integral function $\text{Ei}(x)$ and logarithmic integral li$(x)$. It uses the $n$th derivative formula of Inverse Gamma regularized $\frac{d^n Q^{-1}(a,z)}{dz^n}$. From
we know that Ei$\displaystyle\left(-\lim_{a\to0}Q^{-1}(a,-ay)\right)=y<0$ Here is a Taylor series formula about $z=Q(a,x)$ with Gamma Regularized, the Pochhammer Symbol $(u)_v$, and Kronecker delta $\delta_n$ and $\delta_{n,m}$: $$Q^{-1}(a,z)=\sum_{n=0}^\infty\frac{(z-Q(a,x))^n}{n!}\left(x\delta_n+\left(-\frac{\Gamma(a)e^x}{x^{a-1}}\right)^n\sum_{j_2=0}^n\cdots\sum_{j_n=0}^n(-1)^{\sum\limits_{m=2}^nj_m}\delta_{\sum\limits_{m=2}^n (m-1)j_m,n-1}\Gamma\left(n+\sum_{m=2}^n j_m\right)\prod_{m=2}^n\frac1{j_m !}\left(\frac{a!e^x x^{1-a-m}}{m!}\right)^{j_m} \left(\sum_{k=0}^m(-1)^{m-k}\binom mk (1-a-k)_{m-1}Q(a+k,x)\right)^{j_m}\right)\tag1$$
We have to use the above Taylor series for this branch about $z=\text{Ei}(x)$: $$\text{Ei}^{-1}(z)=x+xe^{-x}(z-\text{Ei}(x))+\frac14e^{-2x} (x-x^2) (z-\text{Ei}(x))^2+e^{-3x}\left(\frac{x^3}3-\frac{2x^2}3+\frac x6\right)(z-\text{Ei}(x))^3+e^{-4x}\left(\frac x{24}-\frac{x^4}4+\frac{3x^3}4-\frac{11x^2}{24}\right)(z-\text{Ei}(x))^4+…$$
shown by this special case. Now we use the formula:
$$\text{Ei}^{-1}(x)=-\lim_{a\to0}Q^{-1}(a,-ax)=-\lim_{a\to0}\sum_{n=0}^\infty\frac{(z-\text{Ei}(x))^n}{n!}\left(-ax\delta_n+\left(-\frac{\Gamma(a)e^{-ax}}{(-ax)^{a-1}}\right)^n\sum_{j_2=0}^n\cdots\sum_{j_n=0}^n(-1)^{\sum\limits_{m=2}^nj_m}\delta_{\sum\limits_{m=2}^2 (m-1)j_m,n-1}\Gamma\left(n+\sum_{m=2}^n j_m\right)\prod_{m=2}^n\frac1{j_m !}\left(\frac{a!e^{-ax} (-ax)^{1-a-m}}{m!}\right)^{j_m} \left(\sum_{k=0}^m(-1)^{m-k}\binom mk (1-a-k)_{m-1}Q(a+k,-ax)\right)^{j_m}\right)$$
Clearly $-ax\delta_n\to0,-\frac{\Gamma(a)e^{-ax}}{(-ax)^{a-1}}\to x$, but what is $$\lim_{a\to0}\prod_{m=2}^n\frac1{j_m !}\left(\frac{a!e^{-ax} (-ax)^{1-a-m}}{m!}\right)^{j_m} \left(\sum_{k=0}^m(-1)^{m-k}\binom mk (1-a-k)_{m-1}Q(a+k,-ax)\right)^{j_m}?$$
It may help that the explicit formula looks like a special case of the explicit series reversion formula
$$$$
Maybe we substituted $Q^{-1}(a,z)\to -\lim\limits_{a\to0}Q^{-1}(a,-az)$ incorrectly. What is the correct way to use $(1)$ to expand $ \text{Ei}^{-1}(z)=-\lim\limits_{a\to0}Q^{-1}(a,-az)$ and what is its radius of convergence?
