A tricky limit involving exponential integrals We define exponential integral according to https://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein
as 
$$\text{Ei}_n(x) = \int_{1}^{\infty} \frac{e^{-xt}}{t^n} dt$$
I'm trying to evaluate the limit of the sequence 
$$ -\frac{i}{ \ln(2)}  \lim_{n\rightarrow \infty} \left[ \text{Ei}_{1- \frac{2i\pi}{\ln(2)}} \left( 2^{-n}   \right)-\text{Ei}_{1+\frac{2i\pi}{\ln(2)}} \left(  2^{-n} \right)  \right] $$ 
Where $n$ is evaluated over positive integers. 
This approaches real number which is very close to the value $\frac{1}{\pi}$ but slightly less than it. (I have confirmed this experimentally).
Work so far:
Attempting to evaluate this is as a limit of a function has been futile, 
Since my inner limit results is an expression that doesn't converge:
$$\lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ \text{Ei}_{1- \frac{2i\pi}{\ln(2)}} \left( 2^{-n}   \right)-\text{Ei}_{1+\frac{2i\pi}{\ln(2)}} \left(  2^{-n} \right)  \right] = \lim_{n \rightarrow \infty, n \in \mathbb{N}} \int_{1}^{\infty}\frac{e^{-2^{-n}t}}{t^{1 - \frac{2i\pi}{\ln{2}}}} dt - \int_{1}^{\infty}\frac{e^{-2^{-n}t}}{t^{1 + \frac{2i\pi}{\ln{2}}}}dt $$ 
= $$ \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[  \int_{1}^{\infty} \frac{e^{-2^{-n}t} \left( t^{1 + \frac{2i\pi}{\ln{2}}} - t^{1 - \frac{2i\pi}{\ln{2}}} \right)}{t^2} dt  \right]$$ 
$$ =\lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[  2i \int_{1}^{\infty} e^{-2^{-n} t} \frac{\sin \frac{2\pi}{\ln 2} \ln t}{t} dt  \right]$$
At this point, "letting $n$ go to infinity" yields the divergent integral:
$$ \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ 2i \int_{1}^{\infty}  \frac{\sin \frac{2\pi}{\ln 2} \ln t}{t} dt \right] $$
Some Developments:
On Suggestion of Sangchul Lee I tried to apply integration by parts to our second to last term (dropping the $2i$ for now).
$$ \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[   \int_{1}^{\infty} e^{-2^{-n} t} \frac{\sin \frac{2\pi}{\ln 2} \ln t}{t} dt  \right]$$
$$ = \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ -\frac{\ln 2}{2\pi }e^{-2^{-n}t} \cos \left( \frac{2\pi}{\ln 2} \ln t \right)_{@[1,\infty]}   - 2^{-n} \frac{\ln 2}{2 \pi} \int_{1}^{\infty} e^{-2^{-n}t}\cos \left( \frac{2\pi}{\ln 2} \ln t \right) dt  \right] $$ 
We can now simplify the first term (taking the evaluation of infinity)
$$ = \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ \frac{\ln 2}{2\pi }e^{-2^{-n}}   - 2^{-n} \frac{\ln 2}{2 \pi} \int_{1}^{\infty} e^{-2^{-n}t}\cos \left( \frac{2\pi}{\ln 2} \ln t \right) dt  \right] $$ 
And we see this becomes:
$$ = \frac{\ln 2}{2\pi } - \frac{\ln 2}{2 \pi } \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ 2^{-n}  \int_{1}^{\infty} e^{-2^{-n}t}\cos \left( \frac{2\pi}{\ln 2} \ln t \right) dt  \right] $$ 
And, after consulting mathematica this reduces to:
$$ \frac{\ln 2}{2\pi } - \frac{\ln 2}{2 \pi } \lim_{n\rightarrow \infty, n\in \mathbb{N}} \left[ \frac{1}{2}  \left(  \Gamma[1 - \frac{2i\pi}{\ln 2} , 2^{-n} ]   +  \Gamma[1 + \frac{2i\pi}{\ln 2} , 2^{-n} ] \right)\right]  $$ 
And that reduces pretty simply to:
$$ \frac{\ln 2}{2\pi } - \frac{\ln 2}{2 \pi }  \left[ \frac{1}{2}  \left(  \Gamma[1 - \frac{2i\pi}{\ln 2} , 0 ]   +  \Gamma[1 + \frac{2i\pi}{\ln 2} , 0 ] \right)\right]  $$ 
And Wolfram is able to verify that this indeed is the result I expect (once I multiply by $\frac{-i(2i)}{\ln 2}$ ) as it is within $10^{-6}$ of $\frac{1}{\pi}$.
My question remains: Is there a better closed form for this term?
 A: Write $\alpha = \frac{2\pi}{\log 2}$ for simplicity. Then we are interested in the limit of the following quantity
\begin{align*}
I_n
:= \frac{\operatorname{Ei}_{1 - i\alpha} (2^{-n}) - \operatorname{Ei}_{1 + i\alpha} (2^{-n})}{i\log 2}.
\end{align*}
Plugging the definition of $\operatorname{Ei}_s$ and taking integration by parts,
\begin{align*}
I_n
&= \frac{1}{i\log 2} \int_{1}^{\infty} \frac{t^{i\alpha} - t^{-i\alpha}}{t}e^{-2^{-n}t} \, \mathrm{d}t
 = \frac{2}{\log 2} \int_{1}^{\infty} \frac{\sin\left( \alpha \log t \right)}{t}e^{-2^{-n}t} \, \mathrm{d}t \\
&\hspace{1em}= \frac{e^{-2^{-n}}}{\pi} - \frac{1}{\pi} \int_{1}^{\infty} \cos\left( \alpha \log t \right) 2^{-n} e^{-2^{-n}t} \, \mathrm{d}t.
\end{align*}
Now apply the substitution $u = 2^{-n}t$ and notice that
$$\cos(\alpha \log t) = \cos(\alpha\log u + \alpha n\log 2) = \cos(\alpha \log u) $$
since $\alpha \log 2 = 2\pi$. Then
\begin{align*}
\lim_{n\to\infty} I_n
&= \lim_{n\to\infty} \left[ \frac{e^{-2^{-n}}}{\pi} - \frac{1}{\pi} \int_{2^{-n}}^{\infty} \cos\left( \alpha \log u \right) e^{-u} \, \mathrm{d}u \right] \\
&= \frac{1}{\pi} - \frac{1}{\pi} \int_{0}^{\infty} \cos\left( \alpha \log u \right) e^{-u} \, \mathrm{d}u
\end{align*}
By recalling that the gamma function is defined as $\Gamma(s) = \int_{0}^{\infty} u^{s-1}e^{-u} \, \mathrm{d}u$, this reduces to
\begin{align*}
\lim_{n\to\infty} I_n
&= \frac{1}{\pi} - \frac{\Gamma(1+i\alpha) + \Gamma(1-i\alpha)}{2\pi}
 = \frac{1}{\pi}\operatorname{Re}\left[ 1 - \Gamma(1+i\alpha) \right].
\end{align*}
But I am skeptical of this having an elementary closed form.
