real analysis - Calculate and provide some justification to get an approximation of $Re(int_0^1(log x)^{-operatorname{li}(x)}dx)$

I was playing with Wolfram Alpha online calculator about possible variations of an integral, the so-called Sophomore's dream. I'm agree that my example is artificious and isn't related with previous nice integral, but I'm curious to know how we can obtain and justify a good approximation of next integral.

Question. Can you provide me a justification for a good approximation (I say a few right digits, four or six) of
$$\Re\left(\int_0^1\frac{1}{(\log x)^{\operatorname{li}(x)}}dx\right),$$
where $\operatorname{li}(x)$ is the logarithmic integral (this MathWorld). Thanks in advance.

I prefer some strategy using analysis, but if you want explain me how get it using a good strategy from the numerical analysis (or combining a numerical method with some facts deduced from the analysis of our integral) it also is welcome.

I add also the plot of our function that you can see using this Wolfram Alpha online calculator

plot Re ((log(x))^(-li(x))), from x=0 to 1

Answer

Solution:

This combination of integrals $$\boxed{A=\int_0^{0.6266}\left(2-e^x+\frac29\sin(6x)\right)\,dx\\B=-\frac9{25}\int_{0.5829}^{0.8685}\cos\left(11x-1.7\right)\,dx\\C=-\frac1{40}\int_{0.8736}^{0.9297}\sin(56x-1.8)\,dx}$$ gives the answer of $$A+B+C=0.3846\cdots$$ which does the job with an error of around $1.3\times10^{-4}$.

For a visualisation see here.

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Strategy:

We split the function into three parts, one between $0$ and around $0.58$ (call this $A$), one between $0.58$ to around $0.87$ (call this $B$) and finally one between $0.87$ to around $0.93$ (call this $C$).

We see that for $A$, the function starts at $1$, rises a little and drops to $0$ at $x=0.58$.

For $B$, the function starts at $0$ at $x=0.58$, reaches a minimum of about $0.36$ and rises to $0$ at $0.87$.

For $C$, the function starts at $0$ at $x=0.87$, rises to around $0.02$ and falls back down to $0$ at $0.93$.

These little rises and falls can be modelled simply by either quadratic or trigonometric functions.

Experimenting with functions in Desmos show that the trig ones are better, and further trial and error gives the coefficients shown in the Solution.