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Computing an integral in Mathematica

POSTED BY: MEK MUS
6 Replies

Looks like :

Integrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*Exp[Exp[t]], {t, -Pi, Pi}] ==
((-1)^(I (n + \[Lambda])) (-Gamma[-I (n + \[Lambda]), -E^\[Pi]] + 
Gamma[-I (n + \[Lambda]), -Cosh[\[Pi]] + Sinh[\[Pi]]]))/(2 \[Pi])

We can check:

f[n_, \[Lambda]_] := 
 NIntegrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*
 Exp[Exp[t]], {t, -Pi, Pi}, Method -> "LocalAdaptive", 
 WorkingPrecision -> 25];
 g[n_, \[Lambda]_] := ((-1)^(
 I (n + \[Lambda])) (-Gamma[-I (n + \[Lambda]), -E^\[Pi]] + 
 Gamma[-I (n + \[Lambda]), -Cosh[\[Pi]] + Sinh[\[Pi]]]))/(
 2 \[Pi]); N[g[1, 1] - f[1, 1], 20]

  (*0.*10^-18 + 0.*10^-19 I*) (* Looks good *)

Mathematica is not a magic box that'll spit out a solution to any problem.

All computer algebra systems, including Mathematica, are limited in their capabilities.

POSTED BY: Mariusz Iwaniuk
POSTED BY: MEK MUS
POSTED BY: Mariusz Iwaniuk

Thanks for the answer. I would like to know if, in calculating the integral above, you expanded the exponential and summarized the terms as I did (by the way, this is the smartest way to do such an integration) or if there is another way to do it that might be relevant in some circumstances. NB( A write error occurs in the integral above now it is corrected : Exp[Exp[It]])

POSTED BY: MEK MUS
POSTED BY: MEK MUS
Posted 4 years ago

What is I[n + \[Lambda]]? I is not a function.

POSTED BY: Rohit Namjoshi
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