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Work out the integral $$\int_0^{2π} e^{\large e^{ix}} \, dx.$$

I am now stuck with this for $2$ days, so please help!

Here is my try:

$$I=\int_0^{2π} e^{\large e^{ix}} dx=\int_0^{2\pi} e^{\large{\cos x+i\sin x}}dx$$ $$=\int_0^{2\pi}e^{cos x}\left(\cos(\sin x) +i\sin(\sin x)\right) dx$$ $$\overset{\large2\pi -x \rightarrow x}=\int_0^{2\pi} e^{\cos x} \left(\cos (\sin x)+i\sin(\sin(2\pi- x)\right)dx$$ $$\Rightarrow 2I=\int_0^{2\pi} e^{{\cos x}} \cdot 2\cos (\sin x)dx$$ $$\text{as}\ \sin(2\pi -x )=-\sin x$$ $$\Rightarrow I=\int_0^{2\pi} e^{\cos x}\cos (\sin x)dx$$ $$=\int_0^{2\pi} \left(e^{\cos x}\cos (\sin x)+e^{\cos (\pi-x)}\cos (\sin x)\right)dx$$ $$=\int_0^{\pi} 2 \left(\frac{e^{\cos x}+e^{-\cos x}}{2}\right)\cos(\sin x)dx$$ $$=2\int_0^\pi \operatorname{cosh}(\cos x)\cos(\sin x)dx $$

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    $\begingroup$ What have you tried in the past two days, and what is your mathematical background? $\endgroup$
    – user88319
    Commented Feb 21, 2019 at 22:04
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    $\begingroup$ I'm not sure I understand. Do you mean you've transformed the integrand to that form and then can't continue? Or that you are trying to get an integral with that integrand? $\endgroup$
    – user88319
    Commented Feb 21, 2019 at 22:12
  • $\begingroup$ I transformed the integrand to the above mentioned formed but couldn't continue $\endgroup$
    – 28ADY0901
    Commented Feb 21, 2019 at 22:14
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    $\begingroup$ Was this an exercise someone posed for you, or is it just a problem you made up yourself? Not every function has a closed-form antiderivative, and I suspect that this one doesn't. (It can be solved via contour integration, but that's a much more advanced technique than is taught in high schools.) $\endgroup$ Commented Feb 21, 2019 at 22:15
  • $\begingroup$ How did you do that transformation? Where did the "$i$" go? $\endgroup$ Commented Feb 21, 2019 at 22:15

4 Answers 4

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One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=\int_0^{2π} e^{\large tf} dx\Rightarrow I'(t)=\int_0^{2\pi} f e^{\large tf}dx$$ But since $(tf)' = \left(te^{ix}\right)'=it e^{ix} =it f$

$$\Rightarrow I'(t)=\frac{1}{it} \int_0^{2\pi} \left(e^{tf}\right)'dx =\frac{e^{tf}}{it}\bigg|_0^{2\pi} $$ Now $e^{ix}$ is periodic with $T=2\pi $ and $e^{2\pi i} = e^0 =1$ so we have:$$I'(t)=\frac{e^t -e^t}{it}=0$$ And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer. $$I(t)=I(0)=\int_0^{2\pi} dx=2\pi$$

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  • $\begingroup$ Thanks it was a good way of doing that but could you also try to proceed using my transformed integrand $\endgroup$
    – 28ADY0901
    Commented Feb 21, 2019 at 22:31
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    $\begingroup$ I couldnt find the post that I remembered but here's a similar one: artofproblemsolving.com/community/… $\endgroup$
    – Zacky
    Commented Feb 21, 2019 at 22:32
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    $\begingroup$ That is the brilliant idea! You basically showed first that the Imaginary part of $\int_0^{2\pi} e^{e^{ix}} dx=0$ with that $2\pi-x =x$ substitution and you were left with the real part. I am not sure though if arriving at $\int_0^\pi \cosh(\cos x)\cos(\sin x) dx$ does simplify things (as a combination of trigonometric function and a hyperbolic one is not so nice), but hey it's nice to know that it equals to $\pi$./// Afterwards you can proceed like in the link I gave on the second link. $\endgroup$
    – Zacky
    Commented Feb 21, 2019 at 22:38
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    $\begingroup$ Thanks a lot I'll try to think and proceed $\endgroup$
    – 28ADY0901
    Commented Feb 21, 2019 at 22:40
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    $\begingroup$ Feynman! My Ghod, that guy could really do integrals! $\endgroup$ Commented Feb 21, 2019 at 23:02
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For any $n\in\mathbb{Z}\setminus\{0\}$ we have $\int_{0}^{2\pi}e^{nix}\,dx = 0.$ It follows that

$$ \int_{0}^{2\pi}e^{e^{ix}}\,dx = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{2\pi}e^{nix}\,dx =\frac{1}{0!}\int_{0}^{2\pi}e^{0ix}\,dx=2\pi.$$

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  • $\begingroup$ This is how I would've done it. $\endgroup$
    – J.G.
    Commented Feb 22, 2019 at 0:00
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Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ \int_{C} \frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $\frac{1}{i}$ and using the resiude theorem then the integral is $2\pi i \; Res_0(f) = 2\pi i \frac{1}{i} = 2\pi$

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  • $\begingroup$ Hmmm seems a bit advance for a person of my skill set $\endgroup$
    – 28ADY0901
    Commented Feb 21, 2019 at 22:25
  • $\begingroup$ Sorry I didn´t knew it, but I hope in the future will help you!! $\endgroup$
    – JoseSquare
    Commented Feb 21, 2019 at 22:27
  • $\begingroup$ Very nicely presented! $\endgroup$
    – Vats Y
    Commented May 5, 2021 at 14:33
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Make the substitution $z = e^{it}$ to transform the integral into

$$\int_\gamma \frac{e^z}{iz}dz = \frac{1}{2\pi i}\int_\gamma \frac{2\pi e^z}{z-0}dz$$

where $\gamma$ is the positively oriented closed circular path of radius $1$ around $0$.

Now by Cauchy's Integral Formula this is $2\pi e^0 = 2\pi.$

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