Sine-like function involving the exponential integral - is this real valued? For $x >0$ and $a > 0$, consider the following:
$$
f(x,a) = \frac{a}{2i} \bigg( e^{i a}\ \mathrm{Ei}(-i a - x) - e^{-i a} \ \mathrm{Ei}(i a - x) \bigg)
$$
Where $\mathrm{Ei}(x) = \int_{x}^{\infty} \frac{e^{-u}}{u} du$ is the exponential integral function.
Is this function real-valued? It seems to me that it is.
Furthermore, is there a name for this function? Or a way to write in terms of other special functions? 
 A: 
Theorem .Let $f$ be a complex valued function then 
$$f(z) - \overline{f(z)} = 2i \,\mathrm{Im}f(z)$$
  $$f(z) + \overline{f(z)} = 2 \,\mathrm{Re}f(z)$$

$\textit{proof}$
Rewrite 
$$f(z) = \mathrm{Re}f(z)+ i \mathrm{Im}f(z)$$
Then 
\begin{align}f(z) - \overline{f(z)} &= \mathrm{Re}f(z)+ i \mathrm{Im}f(z)-\overline{\mathrm{Re}f(z)+ i \mathrm{Im}f(z)}\\
&=\mathrm{Re}f(z)+ i \mathrm{Im}f(z) - \mathrm{Re}f(z)+ i \mathrm{Im}f(z)\\
&=2i \,\mathrm{Im}f(z)
\end{align}
Similarly 
\begin{align}f(z) + \overline{f(z)} &= \mathrm{Re}f(z)+ i \mathrm{Im}f(z)+\overline{\mathrm{Re}f(z)+ i \mathrm{Im}f(z)}\\
&=\mathrm{Re}f(z)+ i \mathrm{Im}f(z) + \mathrm{Re}f(z)- i \mathrm{Im}f(z)\\
&=2\,\mathrm{Re}f(z)
\end{align}


Corollary. If $\overline{f(z)} = f(\bar{z})$ then 
  $$f(z) - f(\bar{z}) = 2i \,\mathrm{Im}f(z)$$
  $$f(z) + f(\bar{z}) = 2 \,\mathrm{Re}f(z)$$


Let $z = x+iy$
\begin{align}e^{iy}\mathrm{Ei}(z) - \overline{e^{iy}\mathrm{Ei}(z)} &= 
e^{iy}\mathrm{Ei}(z) - \overline{e^{iy}}\overline{\mathrm{Ei}(z)} \\
&=e^{iy}\mathrm{Ei}(z) - e^{-iy}\overline{\mathrm{Ei}(z)} \\
&= 2i \, \mathrm{Im} \{ e^{iy}\mathrm{Ei}(z)\}
\end{align}
Hence what is remaining is to prove that 
$$\overline{\mathrm{Ei}(z)} = \mathrm{Ei}(\bar{z})$$
$\textit{proof}$
\begin{align}\overline{\mathrm{Ei}(z)} &= \overline{\int^\infty_1\frac{e^{-zx}}{x}\,dx} \\
 &= \int^\infty_1\overline{\frac{e^{-zx}}{x}}\,dx\\
 &= \int^\infty_1\frac{e^{-\bar{z}x}}{x}\,dx\\
 &= \mathrm{Ei}(\bar{z})
\end{align}
