Fourier transform of squared exponential integral $\operatorname{Ei}^2(-|x|)$ Let $\operatorname{Ei}(x)$ denote the exponential integral:
$$\operatorname{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}tdt.$$
Now consider the function $\operatorname{Ei}(-|x|)$. 
           
Its Fourier transform is
$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\operatorname{Ei}(-|x|)\,e^{i\hspace{.05em}k\hspace{.05em}x}\,dx=-\sqrt{\frac2\pi}\frac{\arctan|k|}{|k|}.$$
I'm interested in a Fourier transform of its square:

$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\operatorname{Ei}^2(-|x|)\,e^{i\hspace{.05em}k\hspace{.05em}x}\,dx=\,?$$

 A: I changed my evaluation slightly, and I was able to get the result in a very simple form.

First notice that $$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx. $$
Then integrating by parts, and assuming for now that $k >0$, 
$$ \begin{align}\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx &= 2 \left(\frac{ \text{Ei}^{2} (-x)\sin (kx)}{k} \Bigg|^{\infty}_{0} - \frac{2}{k} \int_{0}^{\infty}\frac{ \sin(kx) e^{-x} \text{Ei}(-x) }{x} \, dx \right) \\ &= - \frac{4}{k} \int_{0}^{\infty} \frac{ \sin(kx) e^{-x} \text{Ei}(-x) }{x} \, dx \\ &= \frac{4}{k} \int_{0}^{\infty}\frac{ \sin(kx) e^{-x} }{x} \int_{1}^{\infty} \frac{e^{-tx}}{t} \, dt \, dx \\ &= \frac{4}{k} \int_{1}^{\infty} \frac{1}{t} \int_{0}^{\infty} \frac{\sin(kx) e^{-(1+t)x} }{x} \, dx \, dt \\ &= \frac{4}{k} \int_{1}^{\infty} \frac{1}{t} \,  \arctan \left(\frac{k}{1+t} \right) \, dt \tag{1} \\ &= \frac{4}{k} \left( \log(t) \arctan \left(\frac{k}{1+t}  \right) \Bigg|^{\infty}_{1}  + k \int_{1}^{\infty} \frac{\log(t)}{t^{2}+2t+1+k^{2}} \, dt \right)  \\ &= 4 \int_{1}^{\infty} \frac{\log(t)}{t^{2}+2t+1+k^{2}} \, dt  \\ &= - 4 \int_{0}^{1} \frac{\log(u)}{1+2u+(1+k^2)u^{2}} \, du \tag{2} \\ &= - \frac{4}{1+k^{2}} \int_{0}^{1} \frac{\log(u)}{u^{2}+ \frac{2}{1+k^{2}}u+ \frac{1}{1+k^{2}}} \, du \\& = \frac{2i}{k}  \left( \int_{0}^{1} \frac{\log(u)}{u+ \frac{1}{1+ik}} \, du  - \int_{0}^{1} \frac{\log(u)}{u+ \frac{1}{1-ik}} \, du\right) . \end{align}$$
In general, $$ \begin{align} \int \frac{\log(x)}{x+a} \, dx &= \frac{1}{a} \int \frac{\log(x)}{1+ \frac{x}{a}} \, dx \\ &= \log(x) \log \left(1+ \frac{x}{a} \right) - \int \frac{\log(1+ \frac{x}{a})}{x} \, dx \\ &= \log(x) \log \left(1+ \frac{x}{a} \right) + \text{Li}_{2} \left(- \frac{x}{a} \right) +C. \end{align}$$
Using this result we get 
$$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = \frac{2i}{k} \Big( \text{Li}_{2} (-1-ik) - \text{Li}_{2} (-1+ik) \Big) . $$
This appears to be correct for various values of $k$. And the presence of the imaginary unit does not mean that the result is not real valued.
To get the result for the Fourier transform just replace $k$ with $|k|$ and multiply everything by $\frac{1}{\sqrt{2 \pi}}$. 
For $k=0$ the right side of the equation should be interpreted as a limit. 
You will indeed find that $$\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) \, dx= \frac{2 \sqrt{2} \log(2)}{\sqrt{\pi}}.$$
$ $
$(1)$ $\  \int_{0}^{\infty} \frac{\sin bx}{x} e^{-ax} \, dx= \arctan \left(\frac{b}{a} \right) , \, a>0$
$(2)$ Make the substitution $u = \frac{1}{t}$.
EDIT:
Using the fact that the dilogarithm is real-valued on the real axis for $x \le 1$, the answer can be simplified using the Schwarz reflection principle.
$$ \begin{align} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = \frac{2 \sqrt{2}}{ \sqrt{\pi} |k|} \,  \text{Im} \, \text{Li}_{2} (-1+i|k|)  \end{align}$$
