The differentiation of $\operatorname{Li}(t)$ gives $$\frac{1}{\log t}$$
In mathematica using D[LogIntegral[t], t] it confirms this as one would expect.
But I'm having difficulty integrating:
$$\int_a^\infty \frac 1 {\log (t)} \, dt$$
I'm not sure by hand how to do this and it certainly won't compute in mathematica; yet it differentiates $\operatorname{Li}(t).$ I can see that with limit a set to zero there would be a problem with it going to infinity as it approaches the $y$-axis, something like $a=2$ would not present that problem.
A similar question was posed here:
Convergence or Divergence using Limits
but only established that it is divergent and did not explain that it came from $\operatorname{Li}(t)$ and what would be needed to get back.
Further I cannot integrate $$\int_a^\infty \frac{e^{-st}}{\log (t)} \, dt$$
again with $a=0$ this would be a problem I suppose, but $a=2$ should be okay.
Answer
$$\int_a^\infty \frac{dt}{\log t} \tag{$*$}$$ does not converge. $$\log t < t \to \frac{1}{\log t}>\frac{1}{t}$$ and as $$\int_a^\infty \, \frac{dt}{t}$$ diverges, so does the $(*)$
The second integral converges for any $s>0;\;a>1$, but I can't find a closed form and I think It doesn't exist
Hope this helps
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