For appropriately well behaved functions $f(t) $, $g(t)$ the integral of their convolution is the product of their individual integrals.
$$
\int_{-\infty}^{\infty}(f*g)(t) \, dt=\left(\int_{-\infty}^{\infty}f(t) \, dt\right) \left(\int_{-\infty}^{\infty}g(t) \, dt\right)
$$
https://en.m.wikipedia.org/wiki/Convolution#Integration
I need to take the integral of a convolution over a finite range (lower bound effectively $0$ as both $f$ and $g$ are $0$ for negative time).
$$
\int_{-\infty}^{t}(f*g)(\tau) \, d\tau
$$
Am I able to say anything about this? I can always extend the limits with a step function
$$
\int_{-\infty}^{t}(f*g)(\tau) \, d\tau = \int_{-\infty}^{\infty}\Theta(t-\tau) (f*g)(\tau) \, d\tau
$$
Writing out the convolution
$$
\int_{-\infty}^{\infty}\Theta(t-\tau)\int_{-\infty}^{\infty} f(\tau') g(\tau-\tau') \, d\tau' \, d\tau
$$
Sending the step function into the second integral
$$
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta(t-\tau) f(\tau') g(\tau-\tau') \, d\tau' \, d\tau
$$
Perhaps I can rewrite this as the integral over a simple convolution by defining a new function including the step function? I worry that this might break an assumption of the Fubini-Tonelli theorem (https://en.m.wikipedia.org/wiki/Fubini%27s_theorem) which I know is central to being able to rewrite integrals of convolutions as products of integrals. (e.g. the step function will introduce a discontinuity). My math background is not strong enough to fully understand the assumptions of the theorem. I don't know about measurable spaces / functions.
For specificity, $f(t)=-i \Theta(t) e^{-i \alpha t}: \alpha\in\mathbb{R}$ and $g(t)$ is a complex function such that $g(t<0)=0$ and decays exponentially at long times. I also know the Fourier transform of the convolution is the product of the individual Fourier transforms, so I believe that means that they meet all the necessary assumptions.
Any help is greatly appreciated, thanks!
Best Answer
For $g$ bounded supported on $t > 0$ and $t f \in L^1$ then $ \int_{-\infty}^t \int_{-\infty}^\infty f(u)g(v-u)dudv$ converges absolutely thus you can swap the order of integration $= \int_{-\infty}^\infty \int_{-\infty}^\infty f(u) 1_{v< t} g(v-u)dudv=\int_{-\infty}^\infty \int_{-\infty}^\infty f(u) 1_{v< t} g(v-u)dvdu$ $=\int_{-\infty}^\infty \int_{-\infty}^\infty f(u) 1_{w< t-u} g(w)dwdu= \int_{-\infty}^\infty f(u) G(t-u)du$
where $G(t) = \int_{-\infty}^t g(u)du$.
If only $f \in L^1$ but $g(t) = \Theta(t)C e^{iat}$ then $G$ is bounded, replace the integral by a series to obtain something absolutely convergent and to swap $\int,\sum$.