How can I get compute a contour integral of the function $f(z)=\exp(i z)/z$ in the upper circle with center in $(0,0)$ and radius $R$? I want verify that $\lim_{R\to \infty}\int_C f(x)=0$ in mathematica. I try with ContourIntegrate.

  • $\begingroup$ "I try with ContourIntegrate", yeah it's the right tool, how did you try? Please show us your code. $\endgroup$
    – xzczd
    Apr 17 at 7:02
  • $\begingroup$ I am find a command as CountourIntegrate[[Exp[I z]/z, z /[Element] Circle[{0,0},R]] but only with the upper circle $\endgroup$
    – eraldcoil
    Apr 17 at 7:10
  • $\begingroup$ If you need upper circle, then you need to define the upper circle: ContourIntegrate[Exp[I z]/z, z \[Element] Circle[{0, 0}, R, {0, Pi}], Assumptions -> R > 0] Please read the document of CountourIntegrate and Circle more carefully: i.sstatic.net/6i8mrbBM.png $\endgroup$
    – xzczd
    Apr 17 at 7:32
  • $\begingroup$ Why do people want to close this question? While the answer is somewhere in the documentation I for one did not know that Circle took a 3rd argument so it is usefull to have it here? $\endgroup$
    – chris
    Apr 17 at 11:21
  • $\begingroup$ @chris Well I'd say deciding if an answer "is easily found in the documentation" is somewhat subjective. Consider vote to reopen? I myself won't be against the decision if the community decides to reopen it. $\endgroup$
    – xzczd
    Apr 18 at 3:54

2 Answers 2


The following does the job.

ContourIntegrate[Exp[I  z]/z, z \[Element] Circle[{0, 0}, R, {0, Pi}],  Assumptions -> R > 0]

I (\[Pi] - 2 SinIntegral[R])




I am not sure this is what you want and I am not sure the result is correct but...

Let's define the contour

 z = R  Exp[ I  t];

Check it:

ParametricPlot[ReIm[R  Exp[ I  t] /. R -> 1] // Evaluate, {t, 0, Pi}]

enter image description here

Define the function to integrate

f[z_] = Exp[I  z]/z;

Carry out the integration

tt = Integrate[f[z]  D[z, t], {t, 0, Pi}, Assumptions -> R > 0]

Take the limit

Limit[tt, R -> Infinity]

(* 0 *)

  • $\begingroup$ It should be D[z,t] instead of Abs[D[z, t]] in the above (see, for example, Wiki for info). $\endgroup$
    – user64494
    Apr 17 at 7:50
  • $\begingroup$ @user64494 I should I have known since I wrote the original post which lead to ContourIntegrate :-) mathematica.stackexchange.com/q/13734/1089 $\endgroup$
    – chris
    Apr 17 at 8:17

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