Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I'm trying to evaluate $\lim_{e\to 0} \, \frac{i}{e}\int_{\pi }^0 \frac{1-\exp (i e \exp (i \theta ))}{\exp (i \theta )} \, d\theta$ with Mathematica 9.0.1.0 on OS X.

However, I get "Undefined" for this input:

Limit[I/e Integrate[(1 - Exp[I e Exp[I \[Theta]]])/Exp[I \[Theta]], {\[Theta], \[Pi], 0}], e -> 0]

We can see numerically that the correct answer is $-\pi$ by using this code:

With[{e = 0.000001}, I/e NIntegrate[(1 - Exp[I e Exp[I \[Theta]]])/Exp[I \[Theta]], {\[Theta], \[Pi], 0}] ] // Chop

What goes wrong in this process? Is it possible to get the correct result?

share|improve this question
    
Giving an assumption on that epsilon will help. Limit[ I/eps*Integrate[(1 - Exp[I eps Exp[I \[Theta]]])/ Exp[I \[Theta]], {\[Theta], \[Pi], 0}, Assumptions -> 0 < eps < 1/1000], eps -> 0] Out[329]= -\[Pi] –  Daniel Lichtblau Oct 28 '13 at 14:57
add comment

2 Answers

up vote 2 down vote accepted

There seems to be some issue with the definite integral. But you can easily work around it by using indefinite integration then evaluate for the limit of integration (replaced your $\theta$ with $x$ to paste here)

 r = Integrate[(1 - Exp[I e Exp[I x]])/Exp[I x], x];
-((r /. x -> Pi) - (r /. x -> 0));
Limit[I/e*%, e -> 0]

Mathematica graphics

Mathematica 9.01 on windows

One can see something is strange, by doing:

r = Integrate[(1 - Exp[I e Exp[I x]])/Exp[I x], {x, \[Pi], 0}]
Assuming[Re[e] <= 0 && Im[e] == 0, Limit[I/e*r, e -> 0]]

Where did this 1/4096 value come from?

Mathematica graphics

share|improve this answer
add comment
Assuming[s \[Element] Reals && s > 0, 
 Limit[I/s Integrate[(1 - Exp[I s Exp[I t]])/Exp[I t], {t, Pi, 0}], 
  s -> 0]]

yields $-\pi$ (Mathematica 9.0). The integral yields a conditional expression which renders limit undefined without declaring assumptions. There is a difficulty (probably related to periodicity) if you try to bypass by using generate conditions to false. The limit returned is $\pi$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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