Inverse logarithmic integral for x∈(0,1)

Question

When looking for a suitable representation of the logarithmic integral, $li(x)=\int_0^x \frac{dx}{\log{x}}$, one can found many texts for $x>1$, which is understandable because of its relations to $\pi(x)$. I am looking for something different.

While computing some step responses of a pressure driven system, I stumbled upon a differential equation (assume nonnegative values)

$$\frac{dy}{dt}=-b\log{\frac{y}{b}},\quad y(0)\in(0,b),$$

which yields a solution of the form

$$y(t)=b\ li^{-1} \left(-\frac{a}{b}t+li\left(\frac{y(0)}{b}\right)\right), \quad t\in (0,\mathbb{R}).$$

Graphically, the solution is the first part of the logarithmic integral $li(x), x\in(0,1)$ rotated by 90°.

For different values of another, undisclosed, parameter the system yields things like $1-e^{-t}$ and $\tanh(t)$, with a hypergeometric2F1 stuff happening in-between. The overall goal is to estimate the $a$ parameter as well as the undisclosed one ($b$ is known, as it is a part of the experimental setup) from the system step response.

Is there some nice way of representing the inverse logarithmic integral function $li^{-1}(x),\ x\in (0,1)$? Most of what I have seen done to get $li^{-1}(x)$ converges for $x\to\infty$, more general ideas require a finite derivative.

Answer 1

Independently of @Тyma Gaidash's sophisticated and nice solution, I think that I have a rather simple solution.

It starts from the simple remark that the plot of $y=e^{\text{li}(x)}$ is nice and simple for $x \in (0,1)$.

Expanded as a series around $x=1$, we have $$e^{\text{li}(x)-\gamma }=\sum_{n=1}^\infty a_n\,(x-1)^n$$ where the first coefficients are $$\left\{-1,-\frac{1}{2},-\frac{1}{12},-\frac{1}{72},\frac{1}{720}, -\frac{31}{21600},\frac{859}{907200},-\frac{8669}{12700800},\frac {38911}{76204800},-\frac{2703619}{6858432000}\right\}$$ Using power series reversion $$x=1-\sum_{n=1}^\infty b_n\,k^n\qquad \text{where}\qquad k=e^{\text{li}(x)-\gamma }$$ where the first coefficients are $$\left\{1,\frac{1}{2},\frac{5}{12},\frac{31}{72},\frac{361}{720} ,\frac{4537}{7200},\frac{757517}{907200},\frac{2922187}{254016 0},\frac{41478457}{25401600},\frac{3255225203}{1371686400}\right\}$$ This gives a much better approximation of $x_{\text{calculated}}$

Answer 2

If you do not work too close to $x=0$, suing series around $x=1$

$$\text{li}(x)=\log (1-x)+\gamma +\frac{1}{2}(x-1)-\frac{1}{24} (x-1)^2+O\left((x-1)^3\right)$$ Neglecting the last term

$$x\sim 1+2\,W\Bigg(-\frac 12\,\exp\Big(\text{li}(x)-\gamma \Big) \Bigg)$$ where $W(.)$ is Lambert function.

Remember that for large $x$, $\text{li}(x)$ is commonly represented by a polynomial of Lambert functions.

Concerning the quality of the approximation $$\int_0^1 \Big(\log (1-x)+\gamma +\frac{1}{2}(x-1)-\text{li}(x) \Big)^2 \,dx =8.458\times 10^{-4}$$

Some results $$\left( \begin{array}{cc} x_{\text{given}} & x_{\text{calculated}}\\ 0.1 & 0.181602 \\ 0.2 & 0.249592 \\ 0.3 & 0.329146 \\ 0.4 & 0.416256 \\ 0.5 & 0.508398 \\ 0.6 & 0.603865 \\ 0.7 & 0.701474 \\ 0.8 & 0.800397 \\ 0.9 & 0.900045 \\ 1.0 & 1.000000 \\ \end{array} \right)$$

I suppose that we could make better using the next terms; this would lead to some generalized Lambert functions.