# 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.

• Please avoid using  in titles; it takes up an unnecessary amount of space on the front page. Also, your computation of the integral seems to be over the entire circle, not over the arc. – user61527 Jul 17 '14 at 4:06
• Also, there are two ways to connect $1$ and $i$ along the circle – clock- and counterclockwise. – Kaster Jul 17 '14 at 4:09
• then i will have two integrals one along the circle clock and counter clock ? – Rachel Jul 17 '14 at 4:10
• Usually that's the part of the problem – which direction to take to integrate over. But if it's not given, then yes, technically you need to consider both options. – Kaster Jul 17 '14 at 4:12

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.

The integral around the whole circle is $0$ by Cauchy's theorem. The integral around your particular arc is not.
Hint: $e^z$ is analytic everywhere, so you can compute the integral exactly as you would for a real integral.