Fourier transform delayed heaviside function : sinc-like output When I fourier transform (FFT) a (delayed) Step funciton (Heaviside function) I get frequency spectrum that follows a very sinc-like pattern, see figure below (showing the asbolute value of the FFT)
However, when I derive it analytically I should see no such pattern. 
$y (t)=\begin{cases}
0,  & \text{if $ \quad  t <t_d$ } \\[2ex]
A,  & \text{if $ \quad t> t_d$ } 
\end{cases}$
Analytically, a fourier transform would be the fourier transform of the heaviside function delayed by time $t_d$, which holds no sinc pattern.: 
$Y(\omega) = \int_{-\infty}^{\infty} y(t)e^{-i\omega t}dt = \int_{t_d}^{\infty}Ae^{-i\omega t}dt = A[-\frac{e^{-i \omega t}}{i\omega}]_{t_d}^{\infty} = A [-\frac{e^{-i \omega \infty}}{i\omega} + \frac{e^{-i \omega t_d}}{i\omega}] $
--> (Skipping one step)--> $Y(\omega)= A( \pi \delta(\omega) + \frac{1}{i\omega} )e^{-i \omega t_d} $
A different way of looking at the function in the time domain would be to express it as a constant $A$ minus a pulse starting at $t=0$ with pulse width $t_d$ and amplitude $A$. Converting this to the Frequency domain (derivation not shown, though see comment by K.K.McDonald) yields:
$y(t) = A - A\cdot \Pi(\frac{t-t_d/2}{t_d}) \quad$ if $t>0 $, else $\quad y(t)=0$
$Y(\omega) =  A \pi \delta(\omega)- A (2 \frac{sin(\omega t_d)}{\omega} e^{-i\omega t_d/2})$
This does seem to be more consistent with my Matlab FFT output (see figure below). However, both derivations should be correct and consistent if I'm not mistaken. Why wouldn't they be consistent?
 
 A: Found the error in my derivation. My first derivation is not correct. The heaviside cannot be delayed by simply adding a delay term.
SInce $y(t)$ only has a value for $t>t_d$ the fourier transform yields $\int_{t_d}^{\infty}Ae^{-i\omega t}dt$.
The integral $\int_{t_d}^{\infty}Ae^{-i\omega t}dt = A\int_{t_d}^{\infty}cos(\omega t)dt + Ai\int_{t_d}^{\infty}sin(\omega t)dt $ is not defined. This also required for the derivation for the heaviside fourier transform where $\int_{0}^{\infty}Ae^{-i\omega t}dt$ is not defined
Using a briefer notation: Since $\int_{t_d}^{\infty}$ does not hold because of the infinity in the integral, it is possible to rewrite this to:  
$\int_{t_d}^{\infty}Ae^{-i\omega t}dt = \int_{0}^{\infty}Ae^{-i\omega t}dt - \int_{0}^{t_d}Ae^{-i\omega t}dt$ 
The second term ($\int_{0}^{t_d}$) is finite, and therefore defined. The first term is can derived analogous to the fourier transform of the 'normal' (without delay) heaviside function. The first integral on the rhs is the correction term, if you will, which is a pulse in time domain and a sinc in the frequency domain.
A: Notice that step pulse is in time domain: $y(t) = A\Pi(\frac{t-t_d/2}{t_d})$ and $\Pi(\frac{t-a}{b})=\begin{cases}
1,  & \text{if $ \quad a-\frac{b}{2} <t<a + \frac{b}{2}$ } \\[2ex]
0,  & \text{if $ \quad a+\frac{b}{2} <t$ or $ \, t < a - \frac{b}{2}$ }
\end{cases}$
So the fourier transform becomes: $i = \sqrt{-1}$
$$Y(\omega) = \int_{-\infty}^{\infty} y(t)e^{-i\omega t}dt = \int_{0}^{t_d}e^{-i\omega t}dt = \frac{e^{-i\omega t_{d}}-1}{-i\omega}=(e^{\frac{-i\omega t_{d}}{2}}) \frac{e^{\frac{-i\omega t_{d}}{2}}-e^{\frac{i\omega t_{d}}{2}}}{-i\omega}= (e^{\frac{-i\omega t_{d}}{2}}) \frac{-2i \sin(\omega t_d)}{-i\omega}=(e^{\frac{-i\omega t_{d}}{2}}) \frac{2\sin(\omega t_d)}{\omega}$$
We see that $|Y(\omega)| = \frac{2\sin(\omega t_d)}{\omega}$. in other words the envelope of $Y$ is still sinc but its phase have a linear lag with the value $\frac{\omega t_d}{2}$.
Also your answer is wrong. The Fourier transform you've wrote down is for step function:
$$u(t-t_d)\leftrightarrow (\frac{1}{i \omega} + \pi \delta (\omega)) e^{-i\omega t_d}$$
$$\Pi(t-t_d) \leftrightarrow \frac{2\sin(\omega t_d)}{\omega}$$
For the Heaviside function we have:
$$\Pi(t-t_d) = u(t)-u(t-t_d) \leftrightarrow (\frac{1}{i \omega} + \pi \delta (\omega)) -(\frac{1}{i \omega} + \pi \delta (\omega))e^{-i\omega t_d}  = \frac{1}{i\omega}(1-e^{-i\omega t_d}) =e^{\frac{-i\omega t_{d}}{2}} \frac{e^{\frac{i\omega t_{d}}{2}}-e^{\frac{-i\omega t_{d}}{2}}}{iw} = e^{\frac{-i\omega t_{d}}{2}}\frac{2i\sin(\omega t_d)}{i\omega}=e^{\frac{-i\omega t_{d}}{2}}\frac{2\sin(\omega t_d)}{\omega}$$
