Let's assume we have (in vector form)
$$\dot x(t) = A x(t)+L w(t)$$
The solution to this general linear time-invariant non-homogeneous ODE, with null boundary values, can be written as
$$x(t) = \int^t_{0} \exp\{A(t-\tau)\} L w(\tau) \ d \tau$$
The book I'm reading, when solving the original ODE, with a Fourier Transform, writes the solutions as the convolution:
$$x(t) = \int^{\infty}_{-\infty} f(t-\tau)L w(\tau) \ d\tau$$ where $f(t)=\mathcal{F}^{-1}\left[((i\omega)I - A)^{-1}\right]$ and $\mathcal{F}^{-1}$ is the inverse Fourier Transform.
From what I understood, the author states that comparing this solution, to the previous stated solution, we can deduce $f$ to be equal to $\exp\{At\} H(t)$, where $H$ is the Heaviside function.
My doubt is that we have the lower bound of the integral with convolution as zero, due to the presence of the Heaviside function. However, what guarantees that the upper bound of the integral will also be $t$? The convolution is integrated along the whole of real numbers.