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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 $E_1(z):=\int_z^{\infty} \frac{ e^{-t} }{t}$.

asked Jun 6 2011 at 4:22

Chandrasekhar · Answer 1 · 2011-06-06 06:31:25Z UTC

1

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

The proof for this result is given in this link:

answered Jun 6 2011 at 6:31

Chandrasekhar
2,010●1●11●27

	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 2011 at 8:33
	got a derivation from Bender and Orszag, p253, thx – Purpose2011 Jun 6 2011 at 16:26

A properties about expoential integral, 5.1.11 in Abramowitz and Stegun