1
$\begingroup$

I've been looking at the contour integral $\displaystyle \int_C \frac{dz}{e^z - z}$ over various curves. I've checked that the function satisfies the Cauchy-Riemann equations, so it's holomorphic.

The poles are at the points where $e^z - z = 0$, given by the branches of the Lambert W function: $$z = -W_k(-1) = 0.318 \pm 1.33 i,\, 2.06228 \pm 7.58863 i,\, 2.65319 \pm 13.9492 i,\, \ldots$$ Clearly none of these are inside the unit circle.

So if I understand the residue theorem correctly, I would expect the contour integral around the unit circle to be zero.

But when I integrate this numerically in Mathematica I get what looks like $2 \pi$:

NIntegrate[1/(E^(Cos[t] + I Sin[t]) - (Cos[t] + I Sin[t])), {t, 0, 2 \[Pi]}]

6.28319 + 1.11022*10^-16 I

I also tried over a semicircle and got another non-zero answer.

What have I done wrong?

$\endgroup$
  • 1
    $\begingroup$ You've treated $dz$ as $dt$. But $z=e^{it}$, do $dz=ie^{it}\,dt$. $\endgroup$ – Thomas Andrews Aug 16 '16 at 16:09
  • $\begingroup$ So the integral you've computed is actually $\frac{1}{i}\int_C \frac{dz}{z(e^z-z)}$ which has a pole at $z=0$. $\endgroup$ – Thomas Andrews Aug 16 '16 at 16:11
4
$\begingroup$

You are integrating in $dt$ and not in $dz$.

Multiply the integrand by (Cos[t] + I Sin[t])'=(-Sin[t] + I Cos[t]).

Therefore try

NIntegrate[(-Sin[t] + I Cos[t])/(E^(Cos[t] + I Sin[t]) - (Cos[t] + I Sin[t])), {t, 0, 2 [Pi]}]

WolframAlpha says xxx

$\endgroup$
  • $\begingroup$ Of course, that makes sense. $\endgroup$ – Paul Castle Aug 16 '16 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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