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I've tried the following but it didn't work:

Residue[Exp[z - 1/z], {z, 0}]

not even this:

Residue[Exp[1/z], {z, 0}]

Manually, I've computed the residue in the following way:

$$ f(z)=e^{z-\frac{1}{z}}=e^ze^{-\frac{1}{z}}=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}z^{-n}\right)= $$ $$ = \sum_{n=-\infty}^{\infty}\left(\sum_{k=\max(0,n)}^{\infty}\frac{(-1)^{k-n}}{k!(k-n)!}\right)z^n $$

Since the residue is the coefficient of $z^{-1}$, it's equal to:

$$ \sum_{k=0}^{\infty}\frac{(-1)^{k + 1}}{k! (k + 1)!} $$

Going back to Mathematica, I get:

Sum[(-1)^(k + 1)/(k! (k + 1)!), {k, 0, Infinity}]
(* -BesselJ[1, 2] *)

N[-BesselJ[1, 2], 30]
(* -0.576724807756873387202448242269 *)

How can I get that result without manual computations?

Of course, the symbolic result is much preferred.

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You can get the numeric result by using Cauchy's theorem : NIntegrate[ Exp[z - 1/z], {z, 1 + I, -1 + I, -1 - I, 1 - I, 1 + I}]/(2 Pi I). –  b.gatessucks Feb 5 at 21:32

1 Answer 1

up vote 7 down vote accepted

You can use b.gatessucks idea to make it analytically:

f[t_] = Exp[z - 1/z] /. z -> E^(I t) // FullSimplify

then integrate:

1/(2 Pi) Integrate[E^(I t) f[t], {t, 0, 2 Pi}]
(* -BesselJ[1, 2] *)
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