The Fourier transform of a function $f(t)$ using the [1,-1] notation is,

$$\mathcal{F}(\omega)= \int\limits_{-\infty}^{\infty}f(t)e^{-i\omega t}dt.$$

Using Euler's formula, you can think of this as the projecting $f(t)$ onto its cosine components and its sine components.

$$\mathcal{F}(\omega)= \int\limits_{-\infty}^{\infty}f(t)\cos(\omega t)dt-i\int\limits_{-\infty}^{\infty}f(t)\sin(\omega t)dt.$$

Consider a function that is nonzero only in the interval $t_1 < t < t_2$. The Fourier transform in this case is,

$$\mathcal{F}(\omega)= \int\limits_{t_1}^{t_2}f(t)e^{-i\omega t}dt.$$

The following form can be used to define $f(t)$ between $t_1$ and $t_2$. The function $f(t)$ as well as its Fourier transform $\mathcal{F}(\omega)$ are tabulated and plotted.

$f(t)=$ 
where $t_1=$  and $t_2=$  for frequencies $|\omega | < $ .

$t$   $f(t)$

$f(t)$ $t$

The Fourier transform of $f(t)$

The integrals are calculated numerically using a method called Simpson's rule.

$\omega$   $\text{Re}[\mathcal{F}]$   $\text{Im}[\mathcal{F}]$

$\mathcal{F}(\omega )$ $\omega$

Below, the integrand $I=f(t)e^{-i\omega t}$ is plotted in the complex plane from $t_1$ to $t_2$ for a specific frequency. The red point is $\mathcal{F}(\omega )$ at that frequency. Move the slider to see how the integrand changes as the frequency changes.

$\text{Im}[I]$ $\text{Re}[I]$

$\omega=$