Can you justify the existence of a $x_{*}$ solving $$\DeclareMathOperator{\li}{li}\DeclareMathOperator{\erf}{erf}\li(x_{*})=\erf(x_{*})?$$
Here $\li(x)$ is a special function, the so called logarithmic integral, I believe that the formula that you need for $\li(x)$ then is $$\li(x)=\int_\mu^x\frac{dt}{\log t}$$ where $\mu$ as you've read is Soldner's constant, and $$\erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ is the error function.
I would like to see a numerical method computing a good approximation of such $x_*$, since my purpose is learn and refresh mathematicas, I would like to know/rememer what method works to get an approximation of such problem. I don't need a good approximation since I've used an online tool to compute this as $1.9653631831359533312687921...$, but I would like to know an explanation of such method to compute a good approximation for solve an integral equation (I don't take the derivative since I don't know how handle the derivative of $\li(x)$, wich is defined as a P.V.).
Question. Can you justify the existence of a $x_{*}$ solving $$\li(x_{*})=\erf(x_{*}),$$ and give a explanation of a numerical method to compute this?
If you consider the function $$f(x)=\text{li}(x)-\text{erf}(x)$$ one easy way to find its zero is Newton method. Starting from a reasonable guess $x_0$, it will be updated accoreding to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ $$x_{n+1}=x_n-\frac{\text{li}(x_n)-\text{erf}(x_n)}{\frac{1}{\log (x_n)}-\frac{2 e^{-x_n^2}}{\sqrt{\pi }}}$$ Starting using $x_0=2$, Newton method will then generate the following iterates $$x_1=1.9649503996039002033$$ $$x_2=1.9653631233745275939$$ $$x_3=1.9653631831359520789$$ $$x_4=1.9653631831359533313$$ which is the solution for twenty significant figures.
You could have a faster convergence using Halley method (it requires $f''(x)$). Starting again from $x_0=2$, the iterates will be $$x_1=1.9653642096145336397$$ $$x_2=1.9653631831359533313$$ which is the solution for twenty significant figures.
You could even have a faster convergence using higher order methods (Newton is quadratic, Halley is cubic). Let me give you below the first iterate $x_1^{(n)}$ for this methods $$\left( \begin{array}{cc} n & x_1^{(n)} \\ 2 & 1.9649503996039002033 \\ 3 & 1.9653642096145336397 \\ 4 & 1.9653631809125822565 \\ 5 & 1.9653631829431668086 \\ 6 & 1.9653631831480595920 \\ 7 & 1.9653631831361504849 \\ 8 & 1.9653631831359467245 \\ 9 & 1.9653631831359532274 \end{array} \right)$$