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.
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 $m>0$.