HP 50g Romberg Integration

[Gerson W. Barbosa]

(04-04-2014 08:23 PM)Thomas Klemm Wrote:  

(04-03-2014 09:51 PM)Gerson W. Barbosa Wrote:  The actual result is 0.577215664902 ( « 1. Psi NEG» ).

Did you try the substitution \(u=log(x)\) leading to \(\int_{0}^{\infty}-\frac{\log(u)}{e^u}du\)? Or have you started with this and tried to avoid the calculation of an improper integral?

Cheers
Thomas

Hi Thomas,

No, I just used the integral in the form it was presented by Valentin, at the end of an old thread I started:

http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=109056

There are a few mistakes in the post, hopefully mostly linguistic ones. I intended to revised it, but I haven't done it yet. Also, at least one link is not working anymore.

Cheers,

Gerson.