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Dear everyone,

Hi,

How to derive that equation (5.1.11 in Abramowitz and Stegun)

$E_1(z)= - \gamma - \mathrm{ln} z - \sum_{n=1}^{\infty}\frac{ (-1)^n z^n}{n n!} (|arg z|< \pi)$

Thank you very much in advance!!

here the $E_1(z)$ is defined as $E_1(z):=\int_z^{\infty} \frac{ e^{-t} }{t}$.

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1 Answer 1

You can get more information on this at the following pages:

The proof for this result is given in this link:

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Sorry I didn't get the point in the mathworld link, in wikipedia it is said $\mathrm{Ein}(z) = \int_0^z (1-e^{-t})\frac{dt}{t} (1) = \sum_{k=1}^\infty \frac{(-1)^{k+1}z^k}{k\; k!} $ from eq (1) we have $E_{\mathrm{in}}(z) = \mathrm{ln} z - \mathrm{ln} 0 -\int_0^{\infty} \frac{e^{-t}}{t} dt + \int_z^{\infty} \frac{ e^{-t}}{t} dt = \mathrm{ln} z - \mathrm{ln} 0 - \int_0^{\infty} \frac{ e^{-t}}{t} dt + E_1 (z) $, how to get the Euler constant? Thank you for your help –  Purpose2011 Jun 6 '11 at 8:33
    
got a derivation from Bender and Orszag, p253, thx –  Purpose2011 Jun 6 '11 at 16:26

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