My goal is to find the Fourier transform of the product of two functions: The Heaviside step function $\theta(t)$, and $g(t) = A - A\cos(\omega_0 t)$. I want to do this by first Fourier transforming $\theta$ and $g$, and then convolving the results.
I used the Fourier transform convention
$$ F(\omega) = \int_{-\infty}^{+\infty}f(t) e^{i \omega t} dt \qquad \qquad f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{-i \omega t} d\omega $$
to find that the frequency domain representations to be $$ \Theta(\omega) = \frac{i}{\omega} + \pi \delta(\omega), \quad \mathrm{and} \quad G(\omega) = 2\pi A\left(\delta(\omega) -\frac{\delta(\omega-\omega_0)}{2}-\frac{\delta(\omega+\omega_0)}{2}\right). $$
So, now I need to convolve the results. Based on sources such as this, $$ \Theta(\omega)*G(\omega) = \int_{-\infty}^{+\infty} \Theta(x)G(\omega-x)dx $$ should do the trick. The problem is that after repeated attempts, I always get an answer that is a factor of $2\pi$ larger than what I think it should be physically. This makes me suspicious: Is my definition of convolution correct, or do I need to somehow modify it to account for the different Fourier transform convention? (In other words, does the convolution integral change with different Fourier transform conventions, and if so, what should I use for my convention?)
Sidenote: Is this approach valid?
@Ninad Munshi writes: "It's not your definition of convolution that is the problem. Your assumption that the Fourier transform of a product is the convolution of the Fourier transforms is incorrect with this convention"
Note the following line at the bottom of page 98 in Stanford Prof. Brad Osgood's reader: $$ \mathcal{F}(gf) = \mathcal{F}(\mathcal{F}(k * h)^{-}) = k * h = \mathcal{F}g * \mathcal{F}f $$ where $h = \mathcal{F}f$ and $k = \mathcal{F} g$. A proof is provided in the reference linked above. As you can easily see, the Fourier transform of the product is the convolution of the Fourier transforms, which is in contrast to the comment by @Math1000.
Can someone please explain to me either (1) what is meant by @Ninad Munshi and @Math1000, or (2) where the factor of $2\pi$ comes from? I have a guess that the duality operation changing with F.T. convention may be the culprit. Note the added factor of $2\pi$ in the "Duality" section of this page (you'll need to ctrl+f).