I was trying to compute the Fourier Transform of this trapezium shape (starting at the origin), defined by the following piecewise function:
$S(t<0)=0$
$S(0<t\leq T_R)=\frac{H}{T_R}t$
$S(T_R<t\leq T-T_R)=H$
$S(T-T_R<t\leq T)=-\frac{H}{T_R}t+\frac{HT}{T_R}$
$S(t>T)=0$
The Fourier Transform could be defined as (where j is the imaginary number):
$ S(\omega)= (\int_{0}^{T_R}{\frac{H}{T_R}te^{-j\omega t}dt} -\int_{T-T_R}^{T}{\frac{H}{T_R}te^{-j\omega t}dt}) +\int_{T_R}^{T-T_R}{He^{-j\omega t}dt} +\int_{T-T_R}^{T}{\frac{HT}{T_R}e^{-j\omega t}dt} $
To simplify this expression, is it possible to combine the summed integrals in the brackets, since they have the same integrand but different limits.
edit:
To avoid confusion, from this point on, the trapezium height will be defined as h
Starting with the input of @John Wayland Bales in the comments, I have rewritten the function S(t) in terms of Heaviside step functions $(H(t))$ and used that to simplify to only 3 integrals.
$S(t)=\frac{h}{T_R}[H(t)t-H(t-T_R)(t-T_R)-H(t-(T-T_R))(t-(T-T_R))+H(t-T)(t-T)]$
$ \frac{T_R}{h}S(\omega)= \int_{-\infty}^{\infty}{H(t)t*e^{-j\omega t}dt}- \int_{-\infty}^{\infty}{H(t-T_R)(t-T_R)*e^{-j\omega t}dt}- \int_{-\infty}^{\infty}{H(t-(T-T_R))(t-(T-T_R))*e^{-j\omega t}dt}+ \int_{-\infty}^{\infty}{H(t-T)(t-T)*e^{-j\omega t}dt} $
Using the time shift theorem, where appropriate:
$ \frac{T_R}{h}S(\omega)= \int_{0}^{\infty}{t*e^{-j\omega t}dt}- e^{-j\omega T_R}*\int_{0}^{\infty}{(t-T_R)*e^{-j\omega t}dt}- e^{-j\omega (T-T_R)}*\int_{0}^{\infty}{(t-(T-T_R))*e^{-j\omega t}dt}+ e^{-j\omega T}*\int_{0}^{\infty}{(t-T)*e^{-j\omega t}dt} $
$ \frac{T_R}{h}S(\omega)= \int_{0}^{\infty}{t*e^{-j\omega t}dt}- e^{-j\omega T_R}*\int_{0}^{\infty}{t*e^{-j\omega t}dt}+ e^{-j\omega T_R}*\int_{0}^{\infty}{T_R*e^{-j\omega t}dt}- e^{-j\omega (T-T_R)}*\int_{0}^{\infty}{t*e^{-j\omega t}dt}+ e^{-j\omega (T-T_R)}*\int_{0}^{\infty}{T*e^{-j\omega t}dt}- e^{-j\omega (T-T_R)}*\int_{0}^{\infty}{T_R*e^{-j\omega t}dt}+ e^{-j\omega T}*\int_{0}^{\infty}{t*e^{-j\omega t}dt}- e^{-j\omega T}*\int_{0}^{\infty}{T*e^{-j\omega t}dt} $
$ \frac{T_R}{h}S(\omega)=\\ [1+e^{-j\omega T}-e^{-j\omega T_R}-e^{-j\omega (T-T_R)}] \int_{0}^{\infty}{t*e^{-j\omega t}dt}\\ +[e^{-j\omega T_R}-e^{-j\omega (T-T_R)}] \int_{0}^{\infty}{T_R*e^{-j\omega t}dt}\\ +[e^{-j\omega (T-T_R)}-e^{-j\omega T}] \int_{0}^{\infty}{T*e^{-j\omega t}dt} $
Using unit step functions you can re-write the function as follows.
$$ S(t)=\frac{H}{T_R}[t-(t-T_R)U(t-T_R)-(t-(T-T_R))U(t-(T-T_R))+(t-T)U(t-T) ] $$