Exponential integral times exponential Let us define the exponential integral $E_{1}(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \mathrm{d}t$, with $x$ positive real.

Is there any way to simplify the following product?

$$e^{x} E_{1}(x)$$

I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = \Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.
If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.
The only operation in the way to simplify this is integrating by parts the $E_1$:

$$E_1\left( x\right) = \int_x^\infty d\left(e^{-t}lnt\right) + \int_x^\infty e^{-t}lnt\ dt = -e^{-x}lnx + \int_x^\infty e^{-t}lnt\ dt$$

Then:

$$e^xE_1\left( x \right) = -lnx + e^x\int_x^\infty e^{-t}lnt\ dt$$