I want to find

Residue[(x^3 Exp[1/x])/(1 - x^2), {x, 0}]

And when I try to evaluate this I have again:

Residue[(E^(1/x) x^3)/(1 - x^2), {x, 0}]

So the question is: how to find the residue?

  • $\begingroup$ Exp[1/x] has an essential singularity at the origin. $\endgroup$ Commented Nov 1, 2019 at 15:23

3 Answers 3


Not what you are hoping for but is there a pole at x = 0?

Limit[(x^3 Exp[1/x])/(1 - x^2), x -> 0]

(* Indeterminate *)

I think you have an essential singularity which takes all values as x -> 0.

Hope that helps

  • $\begingroup$ Thank you, it is a good idea to use the definition of residue by "limit". $\endgroup$ Commented Nov 1, 2019 at 12:09
  • $\begingroup$ I will think about a type of this point, maybe you are right and it is an essential singularity, I am not sure :) $\endgroup$ Commented Nov 1, 2019 at 12:11
  • $\begingroup$ That is not enough, it can still be a pole. See: en.wikipedia.org/wiki/Essential_singularity You need to check 1/f too. Limit[1/((x^3 Exp[1/x])/(1 - x^2)), x -> 0] (it is also Indeterminate, so indeed essential singularity). $\endgroup$ Commented May 25, 2022 at 12:33

Just because the origin is an essential singularity doesn't mean that the residue does not exist.

The sum of the residues of all of the singularities is 0. Three of the singularities have residues that are easy to compute:

f[x_] := (x^3 Exp[1/x])/(1-x^2)
res1minus = Residue[f[x], {x, -1}]
res1 = Residue[f[x], {x, 1}]
res∞ = Residue[f[x], {x, ∞}]

-(1/(2 E))



So, the residue at 0 is simply:

res0 = - res1minus - res1 - res∞

-(3/2) + 1/(2 E) + E/2

which has the numerical value:

res0 //N


Let's check this using the definition of residues:

r = .5; (* use a radius that only encloses the singularity at 0 *)
1/(2 Pi I) NIntegrate[f[r Exp[I θ]] (r I Exp[I θ]), {θ, 0, 2 π}]

0.0430806 - 1.35532*10^-14 I

  • $\begingroup$ Integrate instead of NIntegrate gives even a symbolic result. $\endgroup$ Commented Nov 1, 2019 at 21:56
  • $\begingroup$ Not here, it does not. Integrate does not provide symbolic result. Also to author: 10^-14 I is fake term, that is just a bug in Mathematica. $\endgroup$ Commented Jan 5, 2022 at 15:35
  • $\begingroup$ In 13.1 it gives 1/2(-3+1/E+E) but you still need to change .5 to 1/2. $\endgroup$ Commented Jul 2, 2022 at 18:39

You can see that your function has an essentially singularity by plot its amplitude on a small circle around the origin. For example

Module[{x = 0.1 Exp[I θ]}, 
 Plot[Abs[(E^(1/x) x^3)/(1 - x^2)], {θ, 0, 2 π}, 
  PlotRange -> All]]

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.