Question

Find the integral

integral from 0 to 2 pi for

e^(e^(i theta )) d theta

where C: \Z\ =1 Oriented clockwise

Answer 1

Answer on Querstion #49111 - Math - Complex Analysis:

Find an integral $\int_0^{2\pi}e^{e^{i\theta}}d\theta$.

Solution

We can't use all of the standard tricks and tools of integration like substitution, trig substitution, parts. I think the best choice is to use Taylor formula. Let's denote $e^{\theta} = t$ and we will have the next $e^t$.

Now we will use Taylor series and we will have the next expression:

$$1 + \frac {t}{1 !} + \frac {t ^ {2}}{2 !} + \frac {t ^ {3}}{3 !} + \dots$$

Now let's integrate our expression and we will have the next:

$$\int_ {0} ^ {2 \pi} e ^ {e ^ {i \theta}} d \theta = \int_ {0} ^ {2 \pi} \left(1 + \frac {e ^ {i \theta}}{1 !} + \frac {e ^ {2 i \theta}}{2 !} + \frac {e ^ {3 i \theta}}{3 !} + \dots\right) d \theta = 1 \Big | _ {0} ^ {2 \pi} = 2 \pi$$

Here we will count integral with power $n = 2k + 1$ but we will show it for $k = 0$

$$\begin{array}{l} \int_ {0} ^ {2 \pi} \frac {e ^ {i \theta}}{1 !} d \theta = \frac {1}{1 !} \left(\int_ {0} ^ {2 \pi} \cos \theta d \theta + i \int_ {0} ^ {2 \pi} \sin \theta d \theta\right) = \frac {1}{1 !} \left(\sin \theta \Big | _ {0} ^ {2 \pi} - \cos \theta \Big | _ {0} ^ {2 \pi}\right) = \\ = \frac {1}{1 !} (\sin 2 \pi - \sin 0 - i (\cos 2 \pi - \cos 0)) = 0 \\ \end{array}$$

Here we will count integral with power $n = 2k$ but we will show it for $k = 1$

$$\begin{array}{l} \int_ {0} ^ {2 \pi} \frac {e ^ {2 i \theta}}{2 !} d \theta = \frac {1}{2 !} \left(\int_ {0} ^ {2 \pi} \cos 2 \theta d \theta + i \int_ {0} ^ {2 \pi} \sin 2 \theta d \theta\right) = \frac {1}{2 !} \frac {1}{2} \left(\sin 2 \theta \Big | _ {0} ^ {2 \pi} - \cos 2 \theta \Big | _ {0} ^ {2 \pi}\right) = \\ = \frac {1}{2 !} \frac {1}{2} (\sin 4 \pi - \sin 0 - i (\cos 4 \pi - \cos 0)) = 0 \\ \end{array}$$

So we don't have any expressions with factorials.

And our integral will be next $\int_0^{2\pi}e^{e^{i\theta}}d\theta = 2\pi$

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