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?

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$\ds{\int_{0}^{\pi}\expo{\ic k\expo{\ic x}}\,\dd x:\ {\large ?}}$

\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.
