Evaluating $J_0(z)=\frac{1}{\pi}\int_{0}^{\pi}\cos\left(z\sin\theta\right)\,d\theta$ I was fascinated by a question I asked before, in which the following equation suddenly appeared.
$$J_0(z)=\frac{1}{\pi}\int_{0}^{\pi}\cos\left(z\sin\theta\right)\,d\theta$$
My question was can this be proved? The anwser was yes. I started with defining:
$$\cos(z) = 1-\frac {z^2}{2!} + \frac{z^4}{4!} ...$$
Further I know that:
$$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$
And by definition:
$$J_0(t)=\sum_{n=0}^{\infty}\frac{(-1)^n t^{2n}}{(n!)^2 2^{2n}}$$
I think I am close but just some missing links. Anyone that can help?
 A: As you said, use Taylor Series for the Cosine:
$$\cos(z\sin\theta) = \sum_{k = 0}^{+\infty} \frac{(-1)^k (z\sin\theta)^{2k}}{(2k)!}$$
Hence the integral becomes
$$\frac{1}{\pi} \sum_{k = 0}^{+\infty} \frac{(-1)^k z^{2k}}{(2k)!}\int_0^{\pi} \sin^{2k}\theta\ \text{d}\theta$$
By a simple integration, we know that 
$$\int_0^{\pi} \sin^{2k}\theta\ \text{d}\theta = \frac{\sqrt{\pi } \Gamma \left(k+\frac{1}{2}\right)}{\Gamma (k+1)}$$
Where $\Gamma$ is the Euler Gamma Function.
Since your $k$ is an integer and positive, we can safely write
$$\Gamma(k+1) = k!$$
and again since $k\in\mathbb{N}$ we also have:
$$\Gamma\left(k + \frac{1}{2}\right) = \frac{(2k-1)!!\sqrt{\pi}}{2^k}$$
Now we have to get rid of the double factorial, but fortunately there is a cute relation which turns it into a simple factorial:
$$(2k - 1)!! = \frac{(2k)!\sqrt{\pi}}{k! 2^k}$$
Now put all together and you get:
$$\frac{1}{\pi}\sum_{k = 0}^{+\infty} \frac{(-1)^k z^{2k}}{(2k)!}\frac{(2k)! \pi}{k! k! 2^{2k}}$$
Which immediately gives you the wanted result.
$$\sum_{k = 0}^{+\infty} \frac{(-1)^k z^{2k}}{(k!)^2 2^{2k}}$$
