# Integral $\int_0^{2π} e^{e^{ix}} dx$

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:

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

• What have you tried in the past two days, and what is your mathematical background? – Strants Feb 21 at 22:04
• 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? – Strants Feb 21 at 22:12
• I transformed the integrand to the above mentioned formed but couldn't continue – Aditya Garg Feb 21 at 22:14
• 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.) – Michael Seifert Feb 21 at 22:15
• How did you do that transformation? Where did the "$i$" go? – Robert Lewis Feb 21 at 22:15

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. $$I(t)=I(0)=\int_0^{2\pi} dx=2\pi$$

• Thanks it was a good way of doing that but could you also try to proceed using my transformed integrand – Aditya Garg Feb 21 at 22:31
• I couldnt find the post that I remembered but here's a similar one: artofproblemsolving.com/community/… – Zacky Feb 21 at 22:32
• 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. – Zacky Feb 21 at 22:38
• Thanks a lot I'll try to think and proceed – Aditya Garg Feb 21 at 22:40
• Feynman! My Ghod, that guy could really do integrals! – Robert Lewis Feb 21 at 23:02

For any $$n\in\mathbb{Z}\setminus\{0\}$$ we have $$\int_{0}^{2\pi}e^{nix}\,dx = 0.$$ It follows that

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

• This is how I would've done it. – J.G. Feb 22 at 0:00

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 $$2\pi i \; Res_0(f) = 2\pi i \frac{1}{i} = 2\pi$$

• Hmmm seems a bit advance for a person of my skill set – Aditya Garg Feb 21 at 22:25
• Sorry I didn´t knew it, but I hope in the future will help you!! – JoseSquare Feb 21 at 22:27