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Computing an integral in Mathematica

Posted 3 months ago
6 Replies
1 Total Likes

Mathematica is unable to compute the following integral

(1/(2*Pi))*Integrate[Exp[-(I[n + \[Lambda]]*t)]*Exp[Exp[It]], {t, -Pi, Pi}]

Unless I ask it how to do. First expand the first exponential then make the integration (simple) and finally perform a summation to collect the terms. The result is then

-((Sin[Pi*(\[Lambda] + n)]*Gamma[-n - \[Lambda], 0, 1])/Pi)

Conclusion: if I am correct where then is the computational power of Mathematica ? I confess that I am not a heavy user of Mathematica.

6 Replies
Posted 3 months ago

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

This is the integral to evaluate

 Integrate[Exp[-(I*(n + \[Lambda])*t)]*Exp[Exp[t]], {t, -Pi, Pi}]

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.

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]])

Yes, I expanded the exponential and summarized the terms as you did.

Sum[Integrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*(Exp[t])^j/
   j!, {t, -Pi, Pi}], {j, 0, Infinity}]

(*-(((-1)^(I n + 
   I \[Lambda]) (Gamma[-I n - I \[Lambda], 0, -E^-\[Pi]] - 
    Gamma[-I (n + \[Lambda]), 0, -E^\[Pi]]))/(2 \[Pi]))*)

Ok, Bye.

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