Problem with the integral in the limit of a piecewise function

I have this piecewise function

$$\psi(x)=\begin{cases} \frac{1}{2} \left[ \left(ln\frac{x}{2}\right)^{2}-\left(Arctanh\sqrt{1-x^{2}} \right)^{2} \right] ,& x\le 1. \\ %1,&\text{if x<0}.\\ %0,&\text{otherwise}. \\ \frac{1}{2} \left[ \left(ln\frac{x}{2}\right)^{2}+\left(Arctan\sqrt{x^{2}-1} \right)^{2} \right] ,& x> 1. \\ \end{cases}$$

the correpsonding graph is

I need to solve this integral

$$\label {eq: seg2017_2 14} F(\omega)=-i \omega e^{\frac{1}{2}i \omega y^{2}}\int_{0}^{\infty} x J_{0}\left( \omega x y \right) e^{i\omega\left( \frac{1}{2}x^{2}-\psi (x) +\phi_{m}(y) \right)}dx$$

Usually Mathematica help me with this kinds of integrals with the commands "NumericQ" and "NIntegrate".

F[y_?NumericQ, w_?NumericQ, g_?NumericQ] := -I*w*
Exp[0.5*I*w*y^2]*
NIntegrate[
x*(BesselJ[0, w*x*y])*
Exp[I*w*(0.5*x^2 - \[Psi][x] + 0.0764808)], {x, 1.1, g},
WorkingPrecision -> 16, MaxRecursion -> 30,
Method -> {GlobalAdaptive, MaxErrorIncreases -> 10000}]


with values parameters $y=0.3$ and $g = 200000$ and $x$ runing from $(0.1,200000)$ instead $(0,\infty)$

The result of this integral is the function $F(\omega)$ that can I plot with

    FUN = LogLogPlot[{Tooltip@Abs@Subscript[F, 1][0.3, w, 200000]}, {w,
0.001, 100}, PlotRange -> {{0.001, 110}, {0.25, 10}},
PlotStyle -> Black, GridLines -> Automatic, ImageSize -> Large,
FrameTicks -> All, PlotPoints -> 40, Frame -> True,
AspectRatio -> 1]


Can solve the integral in the interval $(0.1,0.99)$ or $(1.1,200000)$

This last graph can be made defining the interval of the integral $(1.1,200000)$ to avoid $x=1$, but I need solve the integral in the interval $(0.1,200000)$

The problem is when I start solving this integral in Mathematica, the process crash because the limit of $\psi(x)$ in $x=1$ . I try avoid this using the command "Exclusions->{1}" so Mathematica exclude the values $x=1$ in the integral but not work at all.

I hope to have been clear in defining the problem

• It will lower the threshold for users to enter investigations if you would also provide the Mathematica code for $\psi$. – Henrik Schumacher May 6 '18 at 10:16