Evaluating the integral $\int_{\left|z\right|=\pi}\frac{\left|z\right|e^{-\left|z\right|}}{z}dz$, where the function is not holomorphic

Question

I need to evaluate on the circle $\left|z\right|=\pi$ the integral $$\int_{\left|z\right|=\pi}\frac{\left|z\right|e^{-\left|z\right|}}{z}dz.$$ The function is not holomorphic there. Anyway, I tried to integrate it using polar coordinates and simplyfing the modulo and I got $2\pi e^{-\pi}$ while the result should be $2\pi^2 ie^{-\pi}$. I'm sure is trivial and I overlooked a stupid error. Can anybody tell me where?

Answer 1

Let be $z = \pi e^{i\theta}, \theta\in[0,2\pi]$: $$ \int_{|z|=\pi}\frac{|z|e^{-|z|}}{z}dz = \int_0^{2\pi}\frac{\pi e^{-\pi}}{\pi e^{i\theta}}\pi i e^{i\theta} = 2\pi^2 i e^{-\pi}. $$ But... Cauchy formula can be used: $$ \int_{|z|=\pi}\frac{|z|e^{-|z|}}{z}dz = \int_{|z|=\pi}\frac{\pi e^{-\pi}}{z}dz = 2\pi^2 i e^{-\pi}. $$

Answer 2

If $\gamma(t)=\pi e^{it}$ ($t\in[0,2\pi]$), then\begin{align}\int_{\lvert z\rvert=\pi}\frac{\lvert z\rvert e^{-\lvert z\rvert}}z\,\mathrm dz&=\int_0^{2\pi}\frac{\bigl\lvert\gamma(t)\bigr\rvert e^{-\lvert\gamma(t)\rvert}}{\gamma(t)}\gamma'(t)\,\mathrm dt\\&=\int_0^{2\pi}\pi e^{-\pi}i\,\mathrm dt\\&=2\pi^2ie^{-\pi}.\end{align}