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In my partial answer to this question: Integral involving polylogarithm and an exponential, I arrive at the integral

$$ \int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} ds , ~~~~ (\ast) $$

where $a \in \mathbb{R}$ and $\mathrm{Ei}$ is the standard exponential integral defined by $$ \mathrm{Ei} (z) = \int_{-\infty}^z \frac{e^t}{t} d t , $$ where the Cauchy principal value is taken for $z > 0$. The integral ($\ast$) is very similar to the integrals #28, #30, and #31 on page 198 of https://ia600303.us.archive.org/1/items/jresv73Bn3p191/jresv73Bn3p191_A1b.pdf. I haven't been able to complete the connection. Does anyone know how to express ($\ast$) in terms of known special functions?

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  • $\begingroup$ Mathematica gives the particularly nasty result $$ \frac{1}{2} \left(-2 a^2 \, _3F_3(1,1,1;2,2,2;-a^2)+4 a^2 \, _3F_3(1,1,1;2,2,2;a^2)+{_1F_1}^{(2,0,0)}(0,1,a^2)-{_1F_1}^{ (2,0,0)}(0,1,0)+2 (\log a^2+\gamma) (-\text{Ei}(a^2)+\log a^2+\gamma )\right), $$ where ${_1F_1}^{(2,0,0)}(a,b,x)=\partial_\alpha^2\,{_1F_1}(\alpha,\beta,z)|_{\alpha=a,\beta=b,z=x}$. I would have to spend some time considering how this result is derived. $\endgroup$
    – Aaron Hendrickson
    Commented Feb 3, 2022 at 19:51
  • $\begingroup$ Thank you. I probably should have noted that I ran Mathematica as well and found the same. From experience, I have found that Mathematica may return a complicated result when, in fact, a simpler expression is possible. I am hoping this is also the case here and the hypergeometric functions may be simplified. $\endgroup$
    – o0BlueBeast0o
    Commented Feb 3, 2022 at 19:57
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    $\begingroup$ It does appear that this simplifies. For example, ${_1F_1}^{(2,0,0)}(0,1,0)=0$ and ${_1F_1}^{(2,0,0)}(0,1,a^2)=\log^2(1-a^2)$. The other hypergeometric functions look like they will also simplify to more elementary functions. $\endgroup$
    – Aaron Hendrickson
    Commented Feb 3, 2022 at 20:04
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    $\begingroup$ You also have formulas like this that define your ${_3F_3}$ functions as a limit with $c\to 1$. $\endgroup$
    – Aaron Hendrickson
    Commented Feb 3, 2022 at 20:09
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    $\begingroup$ @AaronHendrickson ${_1 F_1}^{(2, 0, 0)}(0; 1; a^2) = \ln^2(1 - a^2)$ cannot be correct, the derivative doesn't have a singularity at $a = 1$. In terms of the G-function, the integral is $$G_{3, 4}^{2, 3} {\left( a^2 \middle| {1, 1, 1 \atop 1, 1, 0, 0} \right)} - a^2 \hspace {1 px} {_3 F_3}(1, 1, 1; 2, 2, 2; -a^2)$$ (the residues at the multiple poles give the polygamma terms corresponding to the derivative of $(\alpha)_k$). $\endgroup$
    – Maxim
    Commented Feb 11, 2022 at 0:41

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