How to translate Fourier transform table from f to simple omega? The Fourier transform table we are given, lists Time function Vs Fourier transform in terms of f. But I want to have w(simple omega) instead of f.  Is there a method to do the translation easily?
EG: I have rect(t/T) <==> Tsinc(fT). in the table. From that
I need to get (W/pi)*sinc(Wt) <==> rect(w/2W) 


Let
$$X(f) = \int_{-\infty}^{\infty} x(t) \exp(-i 2\pi ft)\,\mathrm dt$$
denote the Fourier transform of $x(t)$ which relationship we denote
as
$$x(t) \leftrightarrow X(f).$$
An alternative definition defines the Fourier transform of
$x(t)$ as $\hat{X}(\omega)$ where 
$$\begin{align*}
\hat{X}(\omega) &= \int_{-\infty}^{\infty} x(t) \exp(-i\omega t)\,\mathrm dt\\
&= \int_{-\infty}^{\infty} x(t) \exp(-i 2\pi (\omega/2\pi) t)\,\mathrm dt\\
&= X(\omega/2\pi).
\end{align*}$$
Thus the "method to do the translation easily" that you are looking
for is as follows.

If we know $X(f)$, the Fourier transform of $x(t)$ with respect to 
  the frequency variable $f$, then we can find $\hat{X}(\omega)$, the
  Fourier transform of $x(t)$ with respect to the radian frequency
  variable $\omega$ simply by substituting $\omega/2\pi$ for $f$
  everywhere in $X(f)$.

As an application to your example, you know the Fourier transform pair
$$\text{rect}(t/T) \leftrightarrow T\text{sinc}(fT)$$ as a transform pair.
Here, the sinc function is defined as
$$\text{sinc}(\alpha) = \begin{cases}
\frac{\sin(\pi\alpha)}{\pi\alpha}, & \alpha \neq 0,\\
1, & \alpha = 0,\end{cases} ~~\text{but see my comment later}$$
Now, the duality theorem of Fourier transforms says
if $x(t) \leftrightarrow X(f)$, then $X(t) \leftrightarrow x(-f)$ which
applies to the transform pair listed above gives

$$T\text{sinc}(tT) \leftrightarrow \text{rect}(f/T)$$

where we have used the fact that $\text{rect}(\cdot)$ is an even function 
of its argument. replacing the constant $T$ by $W/\pi$ and replacing
$f$ by $\omega/2\pi$, the radian frequency transform
pair is

$$\frac{W}{\pi}\text{sinc}(Wt/\pi) \leftrightarrow \text{rect}(\omega/2W)$$

This is not quite the answer you need to arrive at, and so I will let you in on
a secret.  Most people who tend to use radian frequencies to define Fourier
transforms tend to define the sinc function slightly differently from
the definition I gave above. I will use
Sinc for this other definition which is
$$\text{Sinc}(\alpha) = \begin{cases}
\frac{\sin(\alpha)}{\alpha}, & \alpha \neq 0,\\
1, & \alpha = 0.\end{cases} ~~\text{note the difference between sinc and Sinc}$$
Thus we have
$$\frac{W}{\pi}\text{sinc}(Wt/\pi)
= \frac{W}{\pi}\times\frac{\sin(Wt)}{Wt} = \frac{W}{\pi}\text{Sinc}(Wt)$$
and so your transform pair actually is

$$\frac{W}{\pi}\text{Sinc}(Wt) \leftrightarrow \text{rect}(\omega/2W)$$

which is what you have been struggling to show. In other words, you need to 
check the definition of sinc in addition to scaling the frequency variable when
you go from $f$ to $\omega$ in your Fourier transforms.
