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I have a function

f(t) = 2t(H(t)-H(t-2))

and I want to transform it with fourier transformation, but I'm not sure how heaviside acts at least when there are no given limits to the function.

I do know that I'll get there using

$$f_{transform} = \int_{-\infty}^\infty2t(H(t)-H(t-2))e^{-j\omega t}dt $$

but beyond that, I am not sure how to do it by hand.

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    $H(t) - H(t-2) = 1_{t \in [0,2)}$ so you are computing $\int_0^2 2t e^{- i \omega t}dt$ – 2017-02-14
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    What exactly is omega in this statement? – 2017-02-14
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    The frequency variable. You have an (integrable) function $f(t)$ and you are looking at its Fourier transform $F(\omega) = \int_{-\infty}^\infty f(t) e^{-i \omega t}dt$. You'll see later that the Fourier transform is a unitary/orthogonal linear operator, i.e. [its inverse is given by its adjoint](https://en.wikipedia.org/wiki/Fourier_inversion_theorem) $f(t)= \frac{1}{2\pi}\int_{-\infty}^\infty F(\omega) e^{i \omega t}d\omega$ – 2017-02-14

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