I've been totally engaged with exponential integrals for a while. I came across to this limit in my work. I started to calculate the limit as below: currently, I am not sure about my handouts. would you please check it out ? I'll will be happy for any suggestions. ( If someone else has asked this question before or this question could be considered as a duplicated, Inform me, I will delete it immediately ).
$$lim_{x \longrightarrow \infty}e^{-x}E_{1}(x)=?$$
These are my calculations:
$$e^{-x}E_{1}(x)=-Ce^{-x}-e^{-2x}\ln x-e^{-x}\int_{0}^{x}e^{-t}\ln{ t} \ dt\ \ \ [1]$$
Here after, I am feeling uncertain !
$$\int_{0}^{\infty}e^{-t}\ln t dt=-C \ \ \ [2]$$
Where, C is Euler constant.Therefore, we can simplify the limit as:
$$lim_{x \longrightarrow \infty}e^{-x}E_{1}(x)=-lim_{x \longrightarrow \infty}e^{-2x}\ln x$$
Using Lhopital we will get:
$$lim_{x \longrightarrow \infty}e^{-x}E_{1}(x)=\frac{-1}{\infty}=0$$
Reference:
For $[1]$ , $[2]$, look Gradshteyn Eq.$8.212$, and Eq.$4.331$, respectively. Thanks