Closed form of inverse of $(0,1) \ni x\mapsto \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz + \sqrt{-x\log(x)}$

Question

The function $x\mapsto \frac 1 {\sqrt{2\pi}} \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz + \frac 1 {\sqrt{2\pi}} \sqrt{-x\log(x)}$ appeared as cumulative distribution function of a random variable. For an algorithm I need the quantile transformation of this cdf, i.e. the generalized inverse. Is there a way to obtain a closed form of the inverse function, which may contain the inverse of $\Phi (x) = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{z^2} 2 } dz$ or any other special function?

What I tried:

The function $x\mapsto \frac 1 {\sqrt{2\pi}} \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz$ has inverse function $f(y) = \exp (- (\Phi ^{-1}(1 - y))^2)$. Thus the problem reduces to inverte $y \mapsto y + \sqrt{-f(y) \log (f(y))} = y + \Phi^{-1}(1-y) e^{- \frac{- (\Phi^{-1} (1-y))^2}{2}}$, but here I am stuck.

Answer 1

Too long for a comment:

With $f(y)$ the inverse function, the first few coefficients of the series for $f(y)$ about $f(y)=e^{-2}$ are

$$a_n=\left\{\frac{\sqrt{2}}{e},\frac{3}{2},\sqrt{2} e,2 e^2,\frac{25 e^3}{4 \sqrt{2}},\frac{203 e^4}{16},\frac{129 e^5}{2 \sqrt{2}},\frac{1559 e^6}{8},\frac{2743 e^7}{2 \sqrt{2}},\frac{87833 e^8}{16},\frac{1579043 e^9}{32 \sqrt{2}},\frac{31453069 e^{10}}{128},\frac{172014693 e^{11}}{64 \sqrt{2}},\frac{4097307607 e^{12}}{256},\frac{13198808237 e^{13}}{64 \sqrt{2}},\frac{365808959801 e^{14}}{256},\frac{5425581922717 e^{15}}{256 \sqrt{2}},\frac{171492791499467 e^{16}}{1024},\frac{1438592935184979 e^{17}}{512 \sqrt{2}},\frac{51068804745322799 e^{18}}{2048},\frac{956121875317413283 e^{19}}{2048 \sqrt{2}},\frac{37660423557711966481 e^{20}}{8192},\frac{194578848405494636675 e^{21}}{2048 \sqrt{2}},\frac{8420850948835904133145 e^{22}}{8192},\frac{11900089625422782750375 e^{23}}{512 \sqrt{2}},\frac{140317375915480012838165 e^{24}}{512},\frac{27563321019141948054192515 e^{25}}{4096 \sqrt{2}},\frac{1407173174419688575420970525 e^{26}}{16384},\frac{18643581423362891004782922125 e^{27}}{8192 \sqrt{2}},\frac{1024262203390061020445525577235 e^{28}}{32768},\frac{14565616213719545439974663575335 e^{29}}{16384 \sqrt{2}},\ldots\right\}$$ where my convention is $f(y)=f^{-1}(e^{-2})+\sum\frac{a_n(x-f^{-1}(e^{-2}))^n}{n!}$ and $a_1=\sqrt2/e$. A call to FindSequenceFunction (in Mathematica) times out. A nice closed form for these coefficients probably exists, and perhaps that closed form is informative.