I want to solve this integral integral

$$\int_{\gamma} \frac{e^{iz}}{z^2},\quad \gamma(t)=e^{it}\text{ and } 0 \leq t\leq 2\pi$$

My attempt:

$$f(\gamma(t))=\frac{e^{ie^{it}}}{{(e^{it})}^2}={(e^i)^{e^{it}-2t}} \text{ and } \gamma'(t)=ie^{it}$$

So $$\int_{\gamma}\frac{e^{iz}}{z^2}= \int_{o}^{2\pi}{(e^i)^{e^{it}-2t}} \cdot ie^{it}=i\int_{o}^{2\pi}{(e^i)^{e^{it}-t}}$$

from here I don't know how to continue, it's probably something basic, but I can't find it.

  • 1
    $\begingroup$ You could use Cauchy's Integral Formula for Derivatives $\endgroup$
    – Luciano
    Jun 11, 2022 at 21:14

1 Answer 1


This is best solved by the residue theorem, you have an enclosed pole in the contour. Develop the exponential in its Taylor expansion and observe the residue value at $z = 0$.

$$e^{iz} = 1 + (iz) + \frac{(iz)^2}{2!} + ..$$ Hence $$ \frac{e^{iz}}{z^2} = \frac{1}{z^2}+ \frac{i}{z} - \frac{1}{2!} + ..$$ Hence, the residue at zero is just $i$; there is a simple pole at $z= 0$.

The value of the integral is $2\pi i$ times the sum of the enclosed residues so it's $-2\pi$.


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