Integral involving product of dilogarithm and an exponential I am interested in the integral
\begin{equation}
\int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast)
\end{equation}
where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to evaluate the double integral
\begin{equation}
I (a,b) = \int_{\mathbb{R}^2} \arctan^2{ \left( \frac{y+b}{x+a} \right) } e^{- (x^2 + y^2) } d x dy ,
\end{equation}
where $a,b \in \mathbb{R}$. See, for example, this question: Interesting $\arctan$ integral. In fact, I am able to show that
\begin{equation}
I (a,0) =  \frac{\pi}{2} \left( \frac{\pi^2}{6} e^{-a^2} + a^2 \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u \right) . 
\end{equation}
I would very much like to see the trailing term (i.e., ($\ast$)) evaluated in terms of commonly used special functions. I am have not worked with the dilogarithm very much, so I am not sure what might be the best approach to rewrite this. Does anyone here have any suggestions?

This is a partial answer that is too long for a comment.
Let
\begin{align*}
J(a) = \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u .
\end{align*}
Note that
\begin{align*}
\frac{d }{d u} \left( u \mathrm{Li}_2 (u) - u - (1-u)\ln{(1-u)} \right) = \mathrm{Li}_2 (u) . 
\end{align*}
Then, we may integrate-by-parts to find
\begin{align*}
J(a) & = \left( \frac{\pi^2}{6} - 1 \right) e^{- a^2} + a^2 \int_0^1 \left( u \mathrm{Li}_2 (u) - u - (1-u)\ln{(1-u)} \right) e^{-a^2 u} d u \\
& =  \left( \frac{\pi^2}{6} - 1 \right) e^{- a^2} - \frac{1 - (1 + a^2) e^{-a^2}}{a^2} + \frac{1}{a^2} \left( 1 + e^{-a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma - 1 \right) \right) + a^2 \int_0^1 u \mathrm{Li}_2 (u)  e^{-a^2 u} d u \\
& = \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) - \frac{a}{2} J' (a) ,
\end{align*}
where $\gamma$ is the Euler constant and $\mathrm{Ei}$ is the exponential integral. Therefore,
\begin{align*}
J'(a) + \frac{2}{a} J(a) = \frac{2}{a} \left( \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) \right) ,
\end{align*}
and
\begin{align*}
\frac{d }{d a} ( a^2 J (a) ) = 2 a \left( \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) \right) .
\end{align*}
Finally, we must integrate
\begin{align*}
J(a) = \frac{2}{a^2} \int_0^a s \left( \frac{\pi^2}{6} e^{- s^2} + \frac{e^{-s^2}}{s^2} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) \right) d s .
\end{align*}
Simplifying some yields
\begin{align*}
J(a) = \frac{\pi^2}{6} \frac{1}{a^2} \left( 1 - e^{-a^2} \right) + \frac{2}{a^2}  \int_0^a \frac{e^{-s^2}}{s} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) d s .
\end{align*}
As a result,
\begin{align*}
I (a,0) = \pi \left( \frac{\pi^2}{12} + \int_0^a \frac{e^{-s^2}}{s} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) d s \right) . 
\end{align*}
Now I just need to understand this remaining integral. I am hopeful that I can leverage a result from https://ia800303.us.archive.org/1/items/jresv73Bn3p191/jresv73Bn3p191_A1b.pdf. Maybe #25 on page 198?
Update: I was able to push this calculation a little further. If we let $s \mapsto s^2$, then
\begin{align}
I_4 (a,0) = \pi \left( \frac{\pi^2}{12} + \frac{1}{2} \int_0^{a^2} \frac{e^{-s}}{s} \left( \ln{(s)} - \mathrm{Ei}{(s)} + \gamma \right) d s \right) . 
\end{align}
Noting that
\begin{align*}
\frac{e^{-s}}{s} = \frac{d }{d s} \mathrm{Ei} (-s) ,
\end{align*}
we may integrate-by-parts to find
\begin{align*}
I_4 (a,0) = \pi \left( \frac{\pi^2}{12} + \frac{1}{2} \left( \mathrm{Ei} (- a^2) \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) - \int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} d s \right) \right) . 
\end{align*}
So, I guess now the question is
\begin{align*}
\int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} d s = ??? 
\end{align*}
