Problems of HP prime with triple integrals?

[Albert Chan]

(05-31-2020 10:54 AM)Nigel (UK) Wrote:  Strange. If I change variables to u=1/z, then with exact mode ticked and approximate evaluation I get the correct answer to the triple integral at once, after a couple of messages.

With u=1/z, integral is trivial to evaluate, with integration by parts

\(\large \int e^{1\over z}\;dz
= \int e^u\;d({1\over u})
= {e^u \over u} - \int {1 \over u}\;d(e^u)
= {e^u \over u} - Ei(u)
\)