This  level-5 vital article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Low‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles Low This article has been rated as Low-priority on the project's priority scale.

There's something wrong with the final statement on the page. It reads

The left-hand side depends on 'n', but the right-hand side is independent of 'n'. The wolfram's site http://mathworld.wolfram.com/En-Function.html suggests this:

— Preceding unsigned comment added by 70.226.150.26 (talk • contribs) 03:16, 23 October 2005 (UTC)[reply]

Could anybody type a reference to the C++ code for efficient evaluation of Ei(z) with double-complex precision over all the compex plane? dima (talk) 05:32, 13 January 2008 (UTC)[reply]

For the extension of Ei(z) to the complex values of z, we need to put cut somewhere. How about z<0? If no objections, I shall update the definition. dima (talk) 04:45, 5 February 2008 (UTC)[reply]

I'm stuck on the notation "-i0". I assume "-iπ" is negative, imaginary pi...but this article seems to distinguish between negative and positive imaginary zero. What gives? I'm trying to compute Ei, using some old rational approximations for E1.

Hello, Anonynous. Ei is not analytic at zero. It is not even continuous there. There is nothing wrong in the dependence of the limiting value on the way you approach zero.

As for the rational approximations, be careful; most of them are valid only for real values of the argument. You should check the rational approximation before to apply it. For example, plot the modulus of the residual versus real and imaginary parts of the argument. dima (talk) 10:24, 25 September 2008 (UTC)[reply]

I think the anon was simply confused by the notation. I rewrote it in a human-readable form. — Emil J. 16:47, 25 September 2008 (UTC)[reply]

Last night, Arthur Rubin removed a link to an web calculator providing useful free services to users of exponential integrals. Many wikipedia articles on topics that have calculational aspects provide links to web calculators. This was the only calculator link on the article, so the link to that useful service is now gone. Web calcualtors for such relative obscrure functions are rare. Please cite an official policy justiifcation, explain your actions in light of these points, or engage in a conversation as to why you believe this information to be inappropriate. Otherwise I plan to revert the change. Ichbin-dcw (talk) 19:44, 25 May 2010 (UTC)[reply]

The "Convergent series" sections currently says

Integrating the Taylor series for exp(−t)/t, and extracting the logarithmic singularity...

It's not clear to me how "extracting the logarithmic singularity" is done, and where the Euler constant comes from. It seems to me that one needs to first have "lim[z->0, E1(z)+ln(z)] = -gamma" or something equivalent using other methods (like GAMMA'(1) = -gamma via Weiersrass product), then plug into the integral of Taylor expansion of exp(-t)/t and then use the above result to show upper limit of the integral converges to -gamma. Or am I missing something? — Preceding unsigned comment added by Sgia (talk • contribs) 19:42, 5 November 2011 (UTC)[reply]

--Sgia (talk) 19:47, 5 November 2011 (UTC)[reply]

There must be typos in the Allen-Hastings approx formula. b and c vectors should have 5 components to be multiplied by x₄; the former has only four and the latter is malformed. Someone with access to the ref article could correct the formula. 2A01:E34:EF75:CCE0:223:12FF:FE57:5ADD (talk) 22:56, 21 April 2016 (UTC)[reply]

In the Convergent Series section, after the sentence "These alternating series can also be used to give good asymptotic bounds, e.g.", the lower bound is correct, but the upper bound is incorrect. Try graphing both, you can see for yourself it's wrong. — Preceding unsigned comment added by 75.97.50.252 (talk) 18:26, 20 June 2020 (UTC)[reply]

The upper bound is indeed wrong. I tried hard to understand how that could be - I always though that the exponential integral is quickly increasing! Because the exponent increases much quicker than 1/x falls. Check for x=4. There is a plot of Ei, and it's about 20 at x=4, and the proposed upper bound is less than that. I wrote that it was probably derived for small x (but then the bounds can't be meaningful, since 'small' must be specified in that case).

Yaroslav Nikitenko (talk) 13:07, 16 September 2020 (UTC)[reply]

There was a mistake in the above mentioned section. The error as stated was, in fact, larger than the expression used to approximate. I furthermore cleaned up the section a bit and restricted the statement to positive x, as that is the setting given in the source.