Integral of $\cos(\cos x)$ over $[0,2\pi]$

Question

How to compute the following integral?

$$\mathcal{J}_2=\int_{0}^{2\pi}\cos(\cos t)\,dt$$

I'm trying to compute this integral, but I have no idea of how to do it, can someone help me?

Answer 1

Since, by symmetry, $$\int_{0}^{2\pi}\cos(\cos x)dx = 4\int_{0}^{\pi/2}\cos(\cos x)dx$$ we can just use the Taylor series of $\cos x$ and integrate it termwise. What we get is: $$\mathcal{J}_2 = 2\pi\cdot J_0(1)=2\pi\cdot\sum_{m=0}^{+\infty}\frac{(-1)^m}{4^m\,m!^2}=4.8078788612688\ldots$$ (notice that the series converges extremely fast, so it is perfect for numerical purposes)

where $J_0$ is a Bessel function of the first kind.

Answer 2

It obviously cannot be solved by computing the indefinite integral - this is a matter of special functions.

Is this a homework question? What's the subject? You obviously need to delve into the world of Bessel functions, for instance.

http://en.wikipedia.org/wiki/Bessel_function

One thing you can do is: perform suitable translations $u=t+ \alpha$ for $\alpha$ an adequate phase to have it fit into the above list.

With this you should be able to finish your exercise. I don't know if there is a way to obtain a nicer result (it all depends on where you have seen the exercise).

Answer 3

Your integral follows from the identity of the Bessel function

$$ J_n (x) = \frac{1}{2 \pi} \int_{-\pi}^\pi e^{i(n \tau - x \sin(\tau))} \,d\tau $$

with $n=0$.