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It is well-known that we can compute the closed-form of the integrals $$\int_1^{\infty}\frac{\log x}{x^2}dx$$ and $$\int_1^{\infty}\frac{\operatorname{li} (x)}{x^3}dx,$$ where $\operatorname{li} (x)$ is the logarithmic integral. On the other hand we know that was defined in the literature its inverse and denoted as $\operatorname{ali}(x)$, for example see in [1], since by definition (I write this sentence to clarify/emphasize the notation) $$\operatorname{li}(\operatorname{ali}(x))=x,$$ for my purpose I take in this post $x\geq 1$.

As motivation for previous integrals (at least from my point of view) and next questions we've the Theorem 3.3 from [1].

Question 1. (Answered in comments by numerical computations) I wondered what about the definite integral $$\int_1^{\infty}\frac{\operatorname{ali}(x)}{x^3}dx\,?\tag{1}$$ That I am asking is if you can provide a reference or if it isn't in the literature a good calculation of $(1)$.

I've splitted my question in two parts, if you want to add also a contribution for previous Question 1 feel free to do it (I don't know if one can to justify easily that $\int\frac{\operatorname{ali}(x)}{x^3}dx$ is a nonelementary integral).

Question 2. I would like to know a concise calculation to know what about the asymptotic behaviour of $$\int_1^X\frac{\operatorname{ali}(x)}{x^3}dx$$ as $X\to\infty$. Many thanks.

I hope that it is a good question for this site, if the integral is in the literature then feel free to add the reference and I try to search and read the corresponding statement from the literature.

Reference:

[1] J. Arias de Reyna and J. Toulisse, The n-th prime asymptotically, J. Théor. Nombres Bordeaux 25 (2013), 521–555.

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    $\begingroup$ FWIW, Mathematica evaluates this to approximately $1.741$. $\endgroup$
    – YiFan
    Commented Jul 20, 2019 at 1:15
  • $\begingroup$ Many thanks @YiFan $\endgroup$
    – user142929
    Commented Jul 20, 2019 at 6:30
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    $\begingroup$ As noted, this is easy to calculate numerically in Mathematica, if you don't need a closed form: With[{x0 = Quiet[\[FormalX] /. First[Solve[LogIntegral[\[FormalX]] == 1 && 1 < \[FormalX] < 2, \[FormalX]]]]}, NIntegrate[x/(Log[x] LogIntegral[x]^3), {x, x0, Infinity}, WorkingPrecision -> 100]], which yields 1.741935272650614694748847046710991610055712810369420527260197272782982526318585249616925251489695552. $\endgroup$ Commented Aug 4, 2019 at 9:37
  • $\begingroup$ Many thanks for your contribution @J.M.isnotamathematician $\endgroup$
    – user142929
    Commented Aug 4, 2019 at 11:10
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    $\begingroup$ for what it worth, the integral equals $a/2+\int_a^\infty \frac{dt}{2 {\rm li}^2(t)}$, where ${\rm li}(a)=1$. $\endgroup$ Commented Sep 2, 2020 at 18:27

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