integrating $\int_{\gamma}e^zdz$ with $\gamma$ is the arc on the unit circle that unites one with i I am stuck integrating $$\int_{\gamma}e^zdz$$
with $\gamma$ is the arc on the unit circle that unites one with i.
I tried this :
The integrand $\mathrm{e}^z$ is holomorphic for $\vert z \vert \le 1$, therefore the integral vanishes by the Cauchy integral theorem.
Now :
$$ \begin{eqnarray}
  \int_0^{2 \pi} \mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) \mathrm{d} \theta &=&
   \int_0^{2 \pi} \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta))  \right) \mathrm{d} \theta = \left. \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right|_0^{2\pi} = 0
\end{eqnarray}
$$
Indeed:
$$ \begin{eqnarray}
  \mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) &=&  \mathrm{e}^{\cos(\theta)} \cos(\theta) \cos(\sin(\theta)) -  \mathrm{e}^{\cos(\theta)}  \sin(\theta) \sin(\sin(\theta) \\ 
   &=& \mathrm{e}^{\cos(\theta)} \cdot  \frac{\mathrm{d} \sin(\sin(\theta))}{\mathrm{d} \theta} + 
      \frac{\mathrm{d} \mathrm{e}^{\cos(\theta)}}{\mathrm{d} \theta} \cdot \sin(\sin(\theta)) \\
  &=& \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta))  \right)
\end{eqnarray}
$$
Is that correct , please some help to solve this.
 A: Since $e^z$ is entire, that is, holomorphic in all of $\Bbb C$, and
$\dfrac{d(e^z)}{dz} = e^z, \tag{1}$
we have
$\int_\gamma \dfrac{d(e^z)}{dz}dz = \int_\gamma e^z dz \tag{2}$
for any path $\gamma:[a, b] \to \Bbb C$, where $[a, b] \subset \Bbb R$ is a closed interval.  Furthermore,
$\int_\gamma \dfrac{d(e^z)}{dz}dz = e^{\gamma(b)} - e^{\gamma(a)}, \tag{3}$
so if $\gamma$ joins $1$ and $i$, that is, $\gamma(a) = 1$ and $\gamma(b) = i$, then we find from (2) that
$\int_\gamma e^z dz = e^i - e^1, \tag{4}$
right?  And we can take it a little further by noting that
$e^i = \cos 1 + i\sin 1, \tag{5}$
whence
$\int_\gamma e^z dz = e^i - e^1 = (\cos 1 - 1) + i\sin 1; \tag{6}$
since $e^z$ is holomorphic on the entirety of $\Bbb C$, the exact path $\gamma$ doesn't matter; it's the endpoints which influence the integral.
You can read all about this stuff here.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: The integral around the whole circle is $0$ by Cauchy's theorem.  The integral around your particular arc is not.
A: Hint: $e^z$ is analytic everywhere, so you can compute the integral exactly as you would for a real integral.
