A Regularized Beta function limit: $\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a$

Question

The goal is to “generalize” the Exponential Integral $\text{Ei}(x)$ using the Regularized Beta function $\text I_z(a,b)$:

$$f(b,z)=\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a$$

Some clues include:

$$\boxed{f(0,z)\mathop=^\text{lim} \lim_{\begin{matrix} a\to0\\ b\to\infty \end{matrix}}\frac{1-\text I_\frac zb(a,b)}a=-\text{Ei}(-z)}$$

$$f(1,z)=-\ln(z)$$

$$f(2,z)=\frac z2-\ln\left(\frac z2\right)-1$$

$$f(3,z)=-\frac{z^2}{18}+\frac{2z}3-\ln\left(\frac z3\right)-\frac32$$

$$f(4,z)=\frac{z^3}{192}-\frac{3z^2}{32}+\frac{3z}4-\ln(z)+\ln(4)-\frac{11}6$$

Is there a closed form for $$\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a?$$

Please correct me and give me feedback!

Answer 1

Notice there is a Lerch Transcendent and an Incomplete Beta function. After using an integral representation, we get the final answer of:

$$\lim_{a\to0}\frac{1-\text I_\frac zb(a,b)}a=-\text B_{1-\frac zb}(b,0)=\left(1-\frac zb\right)^b\Phi\left(1-\frac zb,1,b\right)$$

Therefore:

$$-\lim_{b\to\infty}\left(1+\frac zb\right)^b\Phi\left(1+\frac zb,1,b\right)=\text{Ei}(z)$$

For $b\in\Bbb N$, the function gives a more accurate series expansion for $\text{Ei}(z)$

Of course, now we can use Inverse Beta Regularized with a limit to invert a branch of the exponential integral without having to use this method. Please correct me and give me feedback!