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is it possible to prove that in the

$$ \lim_{x \to 0^+} E_{1}(x) = \lim_{x \to 0^+} \int_{0}^{\infty} dt \frac{e^{-xt}}{t} = c +1/x+O(x)$$

what would be the constant $c$?

EDIT: this was my fault it should read $ \int_{1}^{\infty} dt \frac{e^{-xt}}{t} = c +1/x+O(x)$

in the limit $ t \to 0 $

Jose Garcia
  • 8,506

1 Answers1

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First, put $u=xt$ to get $$E_1(x)=\int_x^\infty \frac{e^{-u}}{u}\, du\, ,$$ and then integrate by parts. This gives $$E_1(x)=-\log(x) e^{-x} +\int_x^\infty \log(u) e^{-u}\, du$$

Since $\log(u)e^{-u}$ is integrable on $(0,\infty)$, the integral in the right-hand side tends to $c=\int_0^\infty \log(u) e^{-u} du$ as $x\to 0$. Moreover, integrating by parts again we get \begin{eqnarray} \int_x^\infty \log(u)e^{-u}du-c&=&\int_0^x \log(u)e^{-u} du\\ &=&(x\log(x)-x)e^{-x}+\int_0^x (u\log(u)-u)e^{-u}du\\ &=&O(x\log(x))\, , \end{eqnarray} so that $\int_x^\infty \log(u)e^{-u}du=c+O(x\log(x))$. Finally, since $e^{-x}=1+O(x)$ we also have $\log(x)e^{-x}=\log(x)+O(x\log(x))$. Altogether, this gives $$E_1(x)=-\log(x)+c +O(x\log(x))\, ,$$ with $c=\int_0^\infty \log(u) e^{-u} du $.

Etienne
  • 13,636
  • In fact, $\int_0^\infty \log(u)e^{-u}du=-\gamma\equiv -0.577$, where $\gamma$ is the Euler-Mascheroni constant. This integral appear on the wiki page for $\gamma$. – WillG Oct 16 '19 at 03:16