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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.

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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 $$\int_0^\infty e^{-s}\,ds=1.$$

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  • $\begingroup$ Sorry! I still don't get it. $\endgroup$ – widapol Apr 14 '13 at 21:54
  • $\begingroup$ I added some detail. I haven't finished the proof, though. (In part because I am heading for bed, and because I wonder if this is home work?) $\endgroup$ – Harald Hanche-Olsen Apr 14 '13 at 21:58
  • $\begingroup$ Let $t = x+s$ in the integral, the result should be $$xe^xE_1(x)=\int_x^\infty \frac{x}{x+s}e^{-s}\,ds.$$ $\endgroup$ – widapol Apr 14 '13 at 22:04
  • $\begingroup$ No, the lower limit $t=x$ corresponds to $s=0$, surely? $\endgroup$ – Harald Hanche-Olsen Apr 15 '13 at 12:58
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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.

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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.

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  • $\begingroup$ While I appreciate your attempt at humour, Phillycheesesteak reminds me more of this theorem which is different, and should not be confused with, this theorem. $\endgroup$ – Willie Wong Jul 17 '13 at 8:37

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