How to find $xy=F(t)$ from $- \operatorname{Ei}{\left(x y e^{i \pi} \right)} = A t^{2} + B t + C$?

Question

I have the integral equation

$$- \operatorname{Ei}{\left(x y e^{i \pi} \right)} = A t^{2} + B t + C$$

where $\operatorname{Ei}$ is the exponential integral and $A$, $B$, and $C$ are (finite) arbitrary constants.

The approach I have thought of so far is to solve the quadratic equation for $t$ at a given $(x,y)$ point, which comes out as follows.

$$t=\left\{- \frac{B}{2 A} - \frac{\sqrt{- 4 A C - 4 A \operatorname{Ei}{\left(x y e^{i \pi} \right)} + B^{2}}}{2 A}, - \frac{B}{2 A} + \frac{\sqrt{- 4 A C - 4 A \operatorname{Ei}{\left(x y e^{i \pi} \right)} + B^{2}}}{2 A}\right\}$$

But if possible I would like the product $xy$ as a function as a function of $t$. I do not know of any explicit inverse of Ei, but although numerical methods are an acceptable backup plan.

Doe this non-elementary expression have a symbolic expression (e.g. a series expansion) for $xy = F(t)$?

Answer 1

For a given value of $t$, with $z=-xy$, you want to solve for $z$ the equation $$\text{Ei}(z)=a$$ Since, as you wrote, the exponential integral function has no explicit inverse, the problem is numerical.

Assuming that $k$ is large, it is better to consider that we look for the zero of function $$f(z)=\log (\text{Ei}(z))-k$$ For large values of $z$, $$\log (\text{Ei}(z))=z-\log (z)+\frac{1}{z}+O\left(\frac{1}{z^2}\right)$$ So, using the first and second terms, an estimate is given by $$z-\log (z)=k \quad \implies \quad z_0=-W_{-1}\left(-e^{-k}\right)$$ where $W_{-1}(.)$ is the second branch of Lambert function.

For a decent and explicit approximation, perform $\color{red}{\text{one}}$ single iteration of Newton, Halley or Householder methods.

From a computing point of view, this would be quite inexpensive sine the dericatives are simple $$f'(z)=\frac{e^z}{z \text{Ei}(z)}\qquad\qquad f''(z)=-\frac{e^z \left((1-z) \text{Ei}(z)+e^z\right)}{z^2 \text{Ei}(z)^2}$$ Some results $$\left( \begin{array}{cccccc} k & z_0 & \text{Newton}& \text{Halley}& \text{Householder}& \text{solution}\\ 1 & 1.00000 & 1.25148 & 1.30682 & 1.29736 & 1.29890 \\ 2 & 3.14619 & 2.56097 & 2.55958 & 2.56278 & 2.56347 \\ 3 & 4.50524 & 4.03956 & 4.03278 & 4.03283 & 4.03290 \\ 4 & 5.74903 & 5.41302 & 5.40991 & 5.40983 & 5.40983 \\ 5 & 6.93685 & 6.68769 & 6.68645 & 6.68642 & 6.68642 \\ 6 & 8.09072 & 7.89681 & 7.89628 & 7.89627 & 7.89627 \\ 7 & 9.22154 & 9.06369 & 9.06344 & 9.06344 & 9.06344 \\ 8 & 10.3356 & 10.2024 & 10.2023 & 10.2023 & 10.2023 \\ 9 & 11.4368 & 11.3214 & 11.3213 & 11.3213 & 11.3213 \\ 10 & 12.5280 & 12.4259 & 12.4259 & 12.4259 & 12.4259 \end{array} \right)$$