HP 50g Romberg Integration
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04-04-2014, 08:23 PM
Post: #14
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RE: HP 50g Romberg Integration
(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 |
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