How to evaluate $\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx$? I'm trying to evaluate the following integral:
$$
 \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx
$$
where $\zeta >0$ is some positive real number. Since the antiderivative of this function is just in terms of the exponential integral, I decided to go for a different approach.
My attempt
I did the following
$$
 \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx =  \int_0^{\pi} \sum_{n=0}^{\infty}\frac{\left(i \zeta e^{ ix}\right)^n}{n!} \ dx  =   \sum_{n=0}^{\infty}\frac{(i \zeta)^n}{n!} \int_0^{\pi}  e^{nix} \ dx = \sum_{n=0}^{\infty}\frac{(i \zeta)^n}{n! (in)}\left(\underbrace{e^{i\pi n}}_{(-1)^n} -1\right)  = \sum_{n=0}^{\infty}\frac{\zeta^ni^{n-1}}{(n+1)!} \left((-1)^n -1\right)
$$
To then verify if my procedure was correct, I used WolframAlpha to evaluate both sides of the equation for the value $\zeta = 1$. From here I got that
$$
\int_0^{\pi} e^{i e^{ ix}} \ dx = 1.2494... \neq -0.9193... = \sum_{n=0}^{\infty}\frac{i^{n-1}}{(n+1)!} \left((-1)^n -1\right)
$$
I'm not sure where I made my mistake. I think interchanging the integral and the sum is justified since I believe the sum converges absolutely, but now I'm not so sure.
Could anyone tell me where my mistake is? Or alternatively, could anyone tell me how I could evaluate this integral? Thank you!

Edit: Thanks to the comments, I believe that I can simplify the integral to be
$$
\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \pi -2\int_0^\zeta \frac{\sin(t)}{t} \ dt
$$
I'm not sure if the approach I was taking was a good way to show this, but if anyone has any ideas about how I could maybe get here I would greatly appreciate them!
 A: After playing around with the integral for a while, I believe I've found a way to solve the integral and get it in terms of $\text{Si}(\zeta)$.
Let's say we define $F(\zeta)$ as
$$
F(\zeta) := \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx
$$
Here we notice that $F(0) = \int_0^{\pi} 1\ dx = \pi$. Now, from here we can then analyze the derivative of $F$ as follows:
\begin{align}
 F'(\zeta) &= \frac{d}{d\zeta} \int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \int_0^{\pi} \frac{\partial}{\partial \zeta }e^{i \zeta e^{ ix}} \ dx =\int_0^{\pi}e^{i \zeta e^{ ix}}\left(e^{ix}\right)i\ dx \\
&\overset{\color{blue}{u=ix}}{=} \int_0^{i\pi}e^{i \zeta e^u} e^u \ du \overset{\color{blue}{s=e^{u}}}{=}\int_1^{-1}e^{i \zeta s} \ ds = \frac{e^{i \zeta s}}{\zeta i}\Bigg\vert_{s=1}^{s=-1} = \frac{1}{\zeta i}\left(e^{-i\zeta} - e^{i \zeta}\right)\\
&= -\frac{2}{\zeta} \left( \frac{e^{i\zeta}-e^{-i\zeta}}{2i}\right) = -2 \frac{\sin(\zeta)}{\zeta}
\end{align}
recalling that we can put the derivative as a partial inside the integral because of Leibniz's integral rule. On the other hand, by the fundamental theorem of calculus, we can easily see that
$$
\frac{d}{d\zeta}-2\text{Si}(\zeta) =-2 \frac{d}{d\zeta} \int_0^\zeta \frac{\sin(t)}{t} \ dt = -2 \frac{\sin(\zeta)}{\zeta}
$$
And since we've found $2$ functions with the same derivative, we know they must be the same up to a constant, or in other words
$$
 F(\zeta) = -2  \int_0^\zeta \frac{\sin(t)}{t} \ dt + c
$$
But recalling the initial condition we had, we can solve for the value of the constant as follows
$$
F(0) = \pi = \int_0^0 \frac{\sin(t)}{t} \ dt + c = c
$$
and so we get the final result being
$$
\boxed{\int_0^{\pi} e^{i \zeta e^{ ix}} \ dx = \pi -2\int_0^\zeta \frac{\sin(t)}{t} \ dt}
$$

I think that this solution is valid for any $\zeta \in \mathbb{R}$, which means I could generalize the original problem to more than just positive values. I believe I haven't missed any details this time, but if I have please let me know!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_0^{\pi}\expo{\ic\zeta{\large\expo{\ic x}}}\!\!\dd x
\,\right\vert_{\ \zeta\ \in\ \mathbb{R}}}  =
\int_{\large z\ \in\ \expo{\large\ic\,\pars{0,\pi}}}
\expo{\ic\,\zeta z}\,{\dd z \over \ic z}
\\[5mm]= &\
\lim_{\epsilon \to 0^{\large +}}\bracks{%
-\int_{-1}^{-\epsilon}\expo{\ic\,\zeta x}\,{\dd x \over \ic x} -
\int_{\pi}^{0}\exp\pars{\ic\,\zeta\epsilon\expo{\ic\theta}}
\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \ic \epsilon\expo{\ic\theta}}
-\int_{\epsilon}^{1}\expo{\ic\,\zeta x}\,{\dd x \over \ic x}}
\\[5mm] = &\
-\mrm{P.V.}\int_{-1}^{1}\expo{\ic\,\zeta x}\,{\dd x \over \ic x} + \pi =
\pi - \int_{0}^{1}\pars{\expo{\ic\,\zeta x} -
\expo{-\ic\,\zeta x}}\,{\dd x \over \ic x}
\\[5mm] = &\
\pi - 2\int_{0}^{1}{\sin\pars{\zeta x} \over x}\,\dd x =
\pi - 2\,\mrm{sgn}\pars{\zeta}\int_{0}^{\verts{\zeta}}{\sin\pars{x} \over x}\,\dd x
\\[5mm] = &\
\bbx{\large\pi - 2\,\mrm{sgn}\pars{\xi}\,\mrm{Si}\pars{\verts{\zeta}}} \\ &
\end{align}
$\ds{\mrm{Si}}$ is the
Sine Integral Function.
