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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!

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    $\begingroup$ For your solution: $n n!\not = (n+1)!. \int_0^\pi 1dx\not=0.$ $\endgroup$ – Iridescent Aug 20 '20 at 7:19
  • $\begingroup$ @User628759, I think that with the corrections you mention I get the correct result of $$ \pi + \sum_{n=1}^{\infty}\frac{\zeta^ni^{n-1}}{n n!} \left((-1)^n -1\right) $$ Can I simplify the sum on the right even further for a general $\zeta >0$? $\endgroup$ – Robert Lee Aug 20 '20 at 7:27
  • $\begingroup$ You are right now. The resulting sum is not elementary, but it can be expressed by exponential integrals. $\endgroup$ – Iridescent Aug 20 '20 at 7:29
  • $\begingroup$ If I understood correctly, I think I should be able to show that $$ \sum_{n=1}^{\infty}\frac{\zeta^ni^{n-1}}{n n!} \left((-1)^n -1\right) = -2\int_0^\zeta \frac{\sin(t)}{t} \ dt $$ Or is this not true for the values of $\zeta$ I'm working with? $\endgroup$ – Robert Lee Aug 20 '20 at 7:41
  • $\begingroup$ You're right, I'll correct it immediately. Thank you for pointing it out! $\endgroup$ – Robert Lee Aug 20 '20 at 10:52
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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!

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

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