# Limit of integral gives incorrect output

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?

-
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

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 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?

-
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 $\pi$.

-