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.


$$\int_a^\infty \frac{dt}{\log t} \tag{$*$}$$ 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

  • 1
    $\begingroup$ +1 for the answer, and I add that there is no close form for the second integral! $\endgroup$ – Von Neumann Dec 17 '17 at 22:14
  • $\begingroup$ Thanks @Raffaele and Henry Turing. Pity no close form for second integral. $\endgroup$ – onepound Dec 17 '17 at 22:17
  • $\begingroup$ @Raffaele I suppose the remaining question is why can one differentiate Li(x) yet cannot get back from $$\int_a^{\infty } \frac{1}{\log (t)} \, dt$$? $\endgroup$ – onepound Dec 17 '17 at 22:20
  • $\begingroup$ I think this explains one way at least math.stackexchange.com/questions/1185779/… $\endgroup$ – onepound Dec 17 '17 at 22:27
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    $\begingroup$ @onepound The integral above does diverge, hence there is no primitive nor numerical form. Also you cannot go back to a numerical integral, from an analytical function! $\endgroup$ – Von Neumann Dec 17 '17 at 22:34

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

$$\text{li}(+\infty) = \infty$$

$$\text{Li}(+\infty) = \infty$$

  • 1
    $\begingroup$ yes @Henry Turing playing about with mathematica and the link given in my previous comment I did find empirically the upper limit but I did not know why until you kindly explained. I will think about it a bit more in the morning. Thanks for now. $\endgroup$ – onepound Dec 17 '17 at 22:37

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