TITLE: Easiest way to find $\Re\int_{0}^{\pi/2} e^{e^{i\theta}} d\theta$
QUESTION [10 upvotes]: How do we find
$$\Re\left[\int_0^{\large\frac{\pi}{2}} e^{\Large e^{i\theta}}d\theta\right]$$
In the shortest and easiest possible manner?
I cannot think of anything good.

REPLY [8 votes]: Alternatively, using Taylor series of exponential function and Euler's formula we have $$e^{\Large e^{i\theta}}=1+(\cos\theta+i\sin\theta)+\frac{(\cos2\theta+i\sin2\theta)}{2!}+\frac{(\cos3\theta+i\sin3\theta)}{3!}+\cdots$$
We also have 
$$\int_0^{\pi/2}\cos(n\theta)\;d\theta=\frac{\sin\left(\frac{\pi n}{2}\right)}{n}=\begin{cases}\dfrac{(-1)^{n-1}}{2n-1}&,\;\text{for $n$ is odd}\\\\0&,\;\text{for $n$ is even}\end{cases}$$
and series for sine integral, see formula $(9)$
$$\text{Si}\,(x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)(2n-1)!}$$
Therefore
$$\begin{align}\int_0^{\pi/2}\Re\left( e^{\Large e^{i\theta}}\right)d\theta&=\int_0^{\pi/2}\left(1 +\cos\theta+\frac{\cos2\theta}{2!}+\frac{\cos3\theta}{3!}+\cdots\right)d\theta\\&=\frac{\pi}{2}+1-\frac{1}{3\cdot3!}+\frac{1}{5\cdot5!}-\frac{1}{7\cdot7!}+\cdots\\&=\frac{\pi}{2}+\text{Si}\,(1)\end{align}$$