I'm having a difficulty approaching this problem. Any help would be highly appreciated. So I want to show $cosh(z+\frac{1}{z})=\sum b_n(z^n+\frac{1}{z^n})$, where $b_n=\frac{1}{2\pi} \int cosn\theta cosh(2cos \theta)d\theta$.
How can I approach this problem? I have tried writing out $cosh(z+\frac{1}{z})$ as its Laurent series, but not sure how do get the coefficients..
First, recall that
$$\cosh(a+b)=\cosh(a)\cosh(b)+\sinh(a)\sinh(b)$$
And apply Cauchy products, using the well known
$$\cosh(x)=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$$
$$\sinh(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$
This exercise may be solved by means of the Fourier cosine series expansion
$$ f(e^{i\theta})=\sum_{n=0}^\infty a_n \cos(n\theta),$$
with $\displaystyle a_n=\dfrac{\int_{0}^{2\pi} \overline{f(e^{i\theta})}\cos(n\theta)d\theta}{\int_{0}^{2\pi} \cos^2(n\theta)d\theta}$.
Here we recall that for $z=e^{i\theta}$, we have $z^n+\frac{1}{z^n}=2\cos(n\theta)$ ($n \in \mathbb{Z}$).
Thus, using the fact that $\cosh(z+\frac{1}{z})=\cosh(2\cos(\theta))$ and $$\int_{0}^{2\pi} \cos^2(n\theta)d\theta=\pi$$ we realize that $\displaystyle a_n=\frac{1}{2\pi}\int_{0}^{2\pi} 2\cos(n\theta)\cosh(2\cos(\theta))d\theta.$ (essentially the same to be proved but written in polar coordinates).
A good exercise (to get in touch with the complex analysis formalism) is to show that the coefficients $a_n$ computed above may be rewritten as a Cauchy integral formula over the unit disk $|z|=1$, involving the $n-$derivative of $f(z)=\cosh(z+\frac{1}{z})$.