Timeline for The integrals of things looking like $e^{(\frac{a}{z}+\frac{b}{z-c})}$ on closed contours

8 events

when toggle format	what		by	license	comment
Jun 3, 2019 at 9:54	vote	accept	R Mary
Jun 2, 2019 at 22:15	comment	added	Pietro Majer		I'm just saying to write $$\oint_C e^{a/z}e^{b/(z-c)}\Big(\sum_{n=0}^\infty g_mz^m\Big)\,dz=\sum_{n=0}^\infty g_m\Big(\oint_C e^{a/z}e^{b/(z-c)}z^mdz\Big).$$
Jun 2, 2019 at 21:21	comment	added	Carlo Beenakker		I'm actually surprised that even the "simplest" case, the unit-circle contour integral of $e^{1/z}e^{1/(z-2)}$, does not seem to have a closed form expression. In that case the series converges rapidly, in 10 terms the same answer as a direct numerical integration is reached (3.26927).
Jun 2, 2019 at 19:40	comment	added	Carlo Beenakker		thanks, @PietroMajer , but isn't $c_m=2\pi i a^{m+1}/(m+1)!$ ? so that is not just a similar series but the same series, or did I misunderstand you?
Jun 2, 2019 at 19:32	comment	added	Pietro Majer		Computing the value $c_m$ of the integral with $g:=x^m$ for all $m$ would allow a similar power series approach, for $g=\sum_{m=0}^\infty g_mx^m$ would give $\sum_{m=0}^\infty g_mc_m $.
Jun 2, 2019 at 7:12	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 82 characters in body
May 31, 2019 at 19:52	history	edited	Carlo Beenakker	CC BY-SA 4.0	added 107 characters in body
May 31, 2019 at 19:46	history	answered	Carlo Beenakker	CC BY-SA 4.0