I want to demonstrate that the convolution $$(f \ast g)(x)=\int_{-\infty}^{\infty}f(x-\tau)g(\tau) \, d\tau$$ can be written in the form $$(f\ast g)(x)=\frac{1}{2}\int_0^{\infty}g(\tau)\left[f(x-\tau)+f(x+\tau)\right] \, d\tau$$
where $f$ is defined in $x>0$ but extended to $x<0$ by means of an even extension.
From what I can see the sensible place to start is to split the integral up as follows:
$$(f \ast g)(x)=\int_0^{\infty}f(x-\tau)g(\tau) \, d\tau +\int_{-\infty}^0f(x-\tau)g(\tau) \, d\tau$$ and then somehow use that $f(x)=f(-x)$ for $x<0$. How do I actually do this though?