# Limit of Exponential Integral Function

I want to ask how to prove whether the following limit is corrent $$\lim_{x \to 0} \left[ {x{e^x}{E_1}( x )} \right] = 0,$$ with $\displaystyle {E_1}( x ) = \int_x^\infty \frac{e^{ - t}}{t}dt$. I try to run it by Matlab and it seems to be true.

Hint: Substitute $t=x+s$ in the integral. As a result, $$xe^xE_1(x)=\int_0^\infty \frac{x}{x+s}e^{-s}\,ds.$$ Now the integrand goes to $1$ as $x\to\infty$, so a much more likely value of the limit is $$\int_0^\infty e^{-s}\,ds=1.$$
• Let $t = x+s$ in the integral, the result should be $$xe^xE_1(x)=\int_x^\infty \frac{x}{x+s}e^{-s}\,ds.$$ – widapol Apr 14 '13 at 22:04
• No, the lower limit $t=x$ corresponds to $s=0$, surely? – Harald Hanche-Olsen Apr 15 '13 at 12:58
We can use the following results: $${e^x}{E_1}\left( x \right) \le \ln \left( {1 + \frac{1}{x}} \right)$$ and $$\mathop {\lim }\limits_{x \to 0} {\left( {1 + \frac{1}{x}} \right)^x} = 1$$ to arrive at the limit.
We have \begin{align} \frac{1}{2}\log\left(1+\frac{2}{x}\right) < e^x E_1(x) < \log\left(1+\frac{1}{x}\right),\,x>0 \end{align} Multiply everywhere by $x$ and apply the Phillycheesesteak theorem.