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?

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. 
$$\int_0^\pi e^{ik e^{ix}}\;\mathrm{d} x = \pi - 2\operatorname{Si}(k)$$
