I am trying to solve the following Integral, $$I = \oint_{|z| = 1}\frac{1}{e^{e^{\frac{1}{z}}}}dz$$
My approach has been as follows: I took "$\ln$" on both sides: $$\ln(I) = \oint_{|z| = 1}\ln\left(\frac{1}{e^{e^{\frac{1}{z}}}}\right)dz\implies - \oint_{|z|=1}e^{\frac{1}{z}} dz.$$ Then I substitute $\frac{1}{z} = t$, and use Cauchy's Integral formula. Final answer that I get is: $I = e^{2\pi i}$.
Am I correct in doing this. If not,then Why and what should be my approach?
Thanks a ton.
ANupam
No, you are not correct: $\ln(\int f)\not=\int \ln(f))$.
Here it is a simpler approach. After letting $w=1/z$, by the Residue Theorem, $$\begin{align} \oint_{|z| =1} \frac{1}{\exp(e^{1/z})}\,dz &=\oint_{|w |=1} \frac{1}{\exp(e^{w})}\cdot \frac{dw}{w^2}\\ &= 2\pi i\,\operatorname{Res}\left(\frac{1}{w^2 \exp(e^{w})},0\right)\\ &=2\pi i \frac{d}{dw}\left(\exp(-e^{w})\right)_{w=0}\\ &= -\frac{2\pi i}{e}. \end{align}$$