Vanishing integral for spacelike separated points The integral I have is this:
$$
K(x,y) = \int_{\mathbb{R}} \sin(p(\theta) \cdot (x-y)) \ d\theta\ 
$$
where 
$$
p(\theta) = m\begin{pmatrix} \cosh(\theta) \\ \sinh(\theta) \end{pmatrix} \quad (m>0),
$$
and $x$ and $y$ are points in Minkowski space.
For simplicity here, I am considering $\mathbb{R^{1+1}}$ so $1$ time dimension and $1$ space. Also the $\cdot$ is the Minkowski inner product: $x\cdot y = x_0y_0 - x_1y_1$ in this case. 
This integral should vanish for spacelike separated points $x$ and $y$, but I don't know how to show this.
 A: Since $x$ and $y$ are spacelike seperated, the vector $v = x-y$ will be such that
$$
v \cdot v = v_0 v_0 - v_1 v_1 < 0.
$$
Now you can find a Lorentz transformation $\Lambda$ such that
$$
v' = \Lambda v = \left( \begin{matrix} 0 \\ r \end{matrix} \right)
$$
where $r = \sqrt{-v \cdot v}$. To be more explicit there's a $\alpha$ such that
$$
\Lambda = \left( \begin{matrix} \cosh(\alpha) & \sinh(\alpha) \\ \sinh(\alpha) & \cosh(\alpha)  \end{matrix} \right).
$$
Now note that we have $p(\theta) \cdot v = \left( \Lambda p(\theta) \right) \cdot \left( \Lambda v \right) = p(\theta+\alpha) \cdot v'$. So we find
$$
K(x,y) = \int_{\mathbb{R}} \sin(p(\theta) \cdot v ) d\theta = \int_{\mathbb{R}} \sin(p(\theta+\alpha) \cdot v') d\theta = \int_{\mathbb{R}} \sin(p(\theta) \cdot v') d\theta 
$$
where the last equality is acquired by using substitution. So really the only integral you have to calculate is
$$
\int_{\mathbb{R}} \sin(-mr \sinh(\theta)) d\theta =\lim_{m \to \infty} \int_{-m}^m \sin(-mr \sinh(\theta)) d\theta,
$$
and because $\theta \mapsto \sin(-mr \sinh(\theta))$ is an odd function, $\int_{-m}^m \sin(-mr \sinh(\theta)) d\theta$ is zero for every $m>0$.
