Evaluate $\int_{0}^{\pi} e^{ike^{ix}}dx$ Does anyone know how to evaluate 

$\int_{0}^{\pi}e^{ike^{ix}}dx .$ ?

Wolfram Alpha gives the answer as $\pi-2Si(k)$, but the math just gets confusing after I read a bit on the Si function.
If split $e^{ix}$ into $\cos(x)+i\sin(x)$, I just get $\int_{0}^{\pi}e^{-k\sin(x)}e^{ik\cos(x)}dx$ , but I'm not sure what to do from there? Can someone please help, like give me a hint on how to evaluate the integral?
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\pi}\expo{\ic k\expo{\ic x}}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#00f}{\large\int_{0}^{\pi}\expo{\ic k\expo{\ic x}}\,\dd x}
=\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}}
{\expo{\ic kz}}\,{\dd z \over \ic z}
\\[3mm]&=-\ic\braces{-\pp\int_{-1}^{1}{\expo{\ic k x} \over x}\,\dd x
-\lim_{\epsilon \to 0^{+}}\bracks{\int_{\pi}^{0}\expo{\ic k\epsilon\expo{\ic\theta}}\,{\epsilon\expo{\ic\theta}\ic\,\dd \theta \over \epsilon\expo{\ic\theta}}}}
=-2\int_{0}^{k}{\sin\pars{x} \over x}\,\dd x + \pi
\\[3mm]&=\color{#00f}{\large\pi - 2\,{\rm Si}\pars{k}}
\end{align}

where $\ds{{\rm Si}\pars{x}}$ is the Sine Integral Function.
A: The method is:
Let $A(x) = \int e^{a e^{bx}}\;\mathrm{d} x$
$A(x) = \int \frac{e^{au}}{bu} \mathrm{d}u, \quad \text{where } b x=\ln u, \mathrm{d}x = \frac 1 {bu} \mathrm{d}u$
$A(x) = -\frac{\operatorname{Ei}(a u)}{b} + c \quad$ where $\operatorname{Ei}(z)$ is the exponential integral function, and $c$ is an arbitrary constant.
The exponential integral is defined as: $\operatorname{Ei}(z)=-\int_0^z \frac{e^t}{t}\,\mathrm{d}t$
[Which is a "special function" that cannot be represented as elementary functions, so that's all you can do.]
$\therefore A(x) = -\frac{\operatorname{Ei}(a e^{bx})}{b} + c$
Now for the specific case where $a=ik$ and $b=i$: $A(x) = i \operatorname{Ei}(ike^{ix}) + c$
And for the definite integral: $A(\pi) - A(0) = i\left(\operatorname{Ei}(-ik)-\operatorname{Ei}(ik)\right)$
Which you can simplify down on studying the relation between the exponential, sine, and cosine integral functions. 
$$\int_0^\pi e^{ik e^{ix}}\;\mathrm{d} x = \pi - 2\operatorname{Si}(k)$$
