Asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$ 
What is the asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$? As $x\to 1$.

I took the series expansion of the integrand and got $e^{\frac{1}{\log(x)}}\approx 1+\frac{1}{\log(x)}+\frac{1}{2\log(x)}+ \cdot\cdot\cdot$
and then I integrated term by term...
I got...
$\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt \approx x+li(x)+\frac{1}{2}\big(li(x)-\frac{x}{\log(x)}\big)+ \cdot\cdot\cdot$
where $li(x)$ is the logarithmic integral.
How right is this?
 A: Write $\varphi(x) = \exp(1/\log x)$ and note that $\varphi^{-1} = \varphi$. So by substituting $y = \varphi(t)$, or equivalently $t  = \varphi(y)$,
\begin{align*}
I(x)
:= \int_{0}^{x} \exp\left(\frac{1}{\log t}\right) \, \mathrm{d}y
&= \int_{1}^{\varphi(x)} y \varphi'(y) \, \mathrm{d}y \\
&= \left[ y \varphi(y) \right]_{1}^{\varphi(x)} + \int_{\varphi(x)}^{1} \varphi(y) \, \mathrm{d}y \\
&= c + x \varphi(x) - \int_{0}^{\varphi(x)} \varphi(y) \, \mathrm{d}y,
\end{align*}
where
$$c := \int_{0}^{1} \varphi(y) \, \mathrm{d}y \approx 0.27973176.$$
Using the expansion $ \varphi(y) = \sum_{n=0}^{\infty} \frac{1}{n!(\log y)^n} $ which converges uniformly for $0 < y \leq \varphi(x)$, we can perform term-wise integration, obtaining
\begin{align*}
I(x)
&= c + x \varphi(x) - \sum_{n=0}^{\infty} \int_{0}^{\varphi(x)} \frac{1}{n!(\log y)^n} \, \mathrm{d}y \\
&= c  - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \int_{0}^{\varphi(x)} \frac{\mathrm{d}y}{(\log y)^{n+1}} 
\end{align*}
Now by using the integral formula
$$ \int \frac{\mathrm{d}y}{(\log y)^{n+1}}
= \frac{1}{n!} \left( \operatorname{li}(y) - \sum_{k=1}^{n} \frac{(k-1)! y}{(\log y)^k} \right), $$
where $\operatorname{li}(x) = \int_{0}^{x} \frac{\mathrm{d}t}{\log t}$ is the logarithmic integral function, we get
\begin{align*}
I(x)
&= c  - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{n!(n+1)!} \left( \operatorname{li}(\varphi(x)) - \varphi(x) \sum_{k=1}^{n} (k-1)! (\log x)^k \right) \\
&= \boxed{ c - a_0 \operatorname{li}(\varphi(x)) + \varphi(x) \sum_{k=1}^{\infty} (k-1)! a_k (\log x)^k }, \tag{*}
\end{align*}
where
$$ a_k := \sum_{n=0}^{\infty} \frac{1}{(n+k)!(n+k+1)!}. $$
