# Alternative ways to prove that $\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$ without complex analysis?

$$\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$$

I derived this rather incredible result via Cauchy's theorem as I was working through some simple contour integrals. I was wondering if this integral can be solved without complex analysis, and if so how? I can't see any obvious substitutions that can be possible, nor any parametrisations that could simplify the integral through Feynman's method.

• This shows how powerful Cauchy's theorem and Cauchy's integral formulas are. Perhaps someone has access to Risch's algorithm and can tell us whether a closed-form antiderivate exists. I guess this is not the case. – Peter Dec 21 '16 at 22:04
• Feynman works. Hint: add a parameter $t$ such that $I(t)=\int_0^{2\pi}e^{t\cos x}\cos(t\sin x)dx$, then rewrite the whole thing using complex exponentials. – Tom Dec 21 '16 at 22:09
• @Tom Does that mean that an antiderivate exist ? Or is this a nice demonstration of the power of the Feynman-method ? – Peter Dec 21 '16 at 22:11
• The latter. What you'll find out by differentiating the function that is defined as such is that $I'(t) = 0$, therefore $I(1) = I(0) = \int_0^{2\pi}dx = 2\pi$. – Tom Dec 21 '16 at 22:14
• @Tom That sounds beautiful. Are you going to post that as an answer? – Simply Beautiful Art Dec 21 '16 at 22:38

## 2 Answers

Notice first that the integrand is symmetrical about the axis $\theta=\pi$, so we have

$$\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)d\theta = 2\int_0^{\pi}e^{\cos\theta}\cos(\sin\theta)d\theta$$

We define the following function:

\begin{align} I(t) :&= \int_{0}^{\pi}e^{t\cos\theta}\cos(t\sin\theta)d\theta\\ &= \frac{1}{2}\int_{0}^{\pi}\exp\left[\frac{t}{2}(e^{i\theta}+e^{-i\theta})\right]\left[\exp\left(\frac{t}{2}(e^{i\theta}-e^{-i\theta})\right)+\exp\left(-\frac{t}{2}(e^{i\theta}-e^{-i\theta})\right)\right]d\theta\\ &= \frac{1}{2}\int_{0}^{\pi}\left[\exp\left(te^{i\theta}\right)+\exp\left(te^{-i\theta}\right)\right]d\theta \end{align}

We differentiate with respect to $t$ and obtain

\begin{align} I'(t) = \frac{1}{2}\int_0^\pi\left[e^{i\theta}\exp\left(te^{i\theta}\right) + e^{-i\theta}\exp\left(te^{-i\theta}\right)\right]d\theta \end{align}

Let $u:=e^{i\theta}$, then $-idu=e^{i\theta}d\theta$, $u(0) = 1$ and $u(\pi)=-1$. Similarly, let $z:=e^{-i\theta}$, then $idz=e^{-i\theta}d\theta$, $z(0) = 1$ and $z(\pi)=-1$.

Then

\begin{align} I'(t) &= -\frac{i}{2}\int_1^{-1}e^{tu}du + \frac{i}{2}\int_1^{-1}e^{tz}dz\\ &= \frac{i}{2}\left[\int_{-1}^1 e^{tu}du - \int_{-1}^1 e^{tz}dz\right] \\ &= 0 \end{align}

since $u$ and $z$ are dummy variables.

Since $I'(t) = 0$ for any $t$, $I(t) = c$ for some $c$ for all values of $t$.

We easily compute

$$I(0) = \int_0^\pi d\theta = \pi$$

Thus we have the following result:

$$\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)d\theta = 2\pi$$

\begin{align} \int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx &=Re\left( \int_0^{2\pi} e^{i\sin(x)}e^{\cos{x}} dx\right) \\ &= Re \left(\int_0^{2\pi} e^{e^{i x}} dx\right)\\ &= Re \left(\sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} e^{inx} dx\right) \\ &= \sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} \cos(nx) dx \\ &= \int_0^{2\pi} dx = 2\pi \end{align}