0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ It is $= \int_{-\infty}^\infty f(u) G(t-u)du$ where $G(t) = \int_{-\infty}^t g(u)du= (e^{-iat}-1) \Theta(t)$ what do you expect more ? $\endgroup$ – reuns Apr 30 '19 at 19:47
  • $\begingroup$ How did you arrive at that? It looks like you swapped the order of integration. Is that allowed in that situation? I suppose that's what my whole question boils down to. I'm not sure if my "modified" situation still meets all the assumptions necessary to invoke the Fubini-Tonelli theorem $\endgroup$ – bRost03 Apr 30 '19 at 20:28
  • $\begingroup$ Perfect, thanks! If you want to put that in an answer I'll happily accept it. $\endgroup$ – bRost03 Apr 30 '19 at 20:37
1
$\begingroup$

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 $\int,\sum$.

$\endgroup$
  • $\begingroup$ Does this mean you are modifying your comment above? I cannot simply swap the order because $g$ is only bounded but not convergent? Can you say something about "replace the integral by a series"? What series, which integral? Thanks for your help! $\endgroup$ – bRost03 Apr 30 '19 at 21:08
  • $\begingroup$ You said $f$ decays exponentially so $t f \in L^1$. If only $f \in L^1$ then you need to keep grouped the integration of $g$ on each period to use that $G$ is bounded while $\int_{-\infty}^t |g(u)|du$ is not $\endgroup$ – reuns Apr 30 '19 at 21:21
  • $\begingroup$ I'm not sure I really understand all that, I don't know what $L^1$ is. But $|g(u)| = \Theta(u)$, so for finite $t$, $\int_{-\infty}^t |g(u)|du$ is also finite. I'm not exactly sure what "bounded" means in this context. I'm not going to mark this as accepted yet because it's not as simple as you'd said above and I'm still not sure what to do. Again, thanks for taking the time to help me out! $\endgroup$ – bRost03 Apr 30 '19 at 21:30
  • $\begingroup$ $f \in L^1$ means $\int_{-\infty}^\infty |f(t)|dt$ converges. Bounded means $\forall t, |G(t)| \le c$. The main idea is that if a double series $\sum_{n,m} a_{n,m}$ converges absolutely then the order of summation doesn't matter, otherwise the order of summation counts, for integrals the idea is the same. $\endgroup$ – reuns Apr 30 '19 at 21:31
  • $\begingroup$ Okay, thanks. If you have the time, I'd appreciate some info on how to do things using the series method you mentioned. If not I understand. $\endgroup$ – bRost03 Apr 30 '19 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.