Integrating definite integral $\int_a^\infty \frac{1}{\log (t)} \, dt$ and then $\int_a^\infty \frac{e^{-st}}{\log (t)} \, dt$ 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.
 A: $$\int_a^\infty \frac{dt}{\log t} \tag{$*$}$$ 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
A: Just to add some details.
The Special Function called "Logarithmic Integral" is defined as follows:
$$\text{li}(x) = \int_0^x \frac{dt}{\ln(t)}$$
By asymptotic methods, we can find its series expansions as
$$\text{li}(z) = \gamma + \ln(-\ln(z)) + \sum_{k = 1}^{+\infty} \frac{(\ln(z))^k}{k\cdot k!}$$
Or, due to Ramanujan, we also have a more rapidly convergent series:
$$\text{li}(z) = \gamma +\ln(\ln(z))  + \sqrt{z}\sum_{k = 1}^{+\infty} \frac{(-1)^{k-1} (\ln(z))^k}{k! 2^{k-1}} \sum_{j = 0}^{\text{floor}{\frac{k-1}{2}}} \frac{1}{2j+1}$$
Nonetheless, the logarithmic integral does diverge at $x = \infty$ hence the integral from $a$ to $\infty$ does diverge.
If instead of the infinite you have $x$, then you can define the so called Offsed Log Integral:
$$\text{Li}(x) = \int_a^x \frac{dt}{\ln(t)}$$
Defined as $$\text{Li}(x) = \text{li}(a) - \text{li}(x)$$
But it still holds that
$$\text{li}(+\infty) = \infty$$
$$\text{Li}(+\infty) = \infty$$
