Fourier transform properties I have to find the Fourier transform of $$ x(t)= \frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) $$

First I applied traslation property , so $$ F[x(t-T)] = X(f) e^{-i 2 \pi f T} $$

after I applied time scaling property , and I obtained 
$$ F[x(\frac{t}{T})] = |T| X(Tf) e^{-i 2 \pi f T} $$

Now I calculate $$ \int_{T}^{+\infty } \frac{1}{T}Te^{-\frac{t-T}{T}}u(t-T) e^{-i 2 \pi f t } $$ that’s results $$ \frac{T}{1+ 2 i\pi f t } $$

So I obtained that the final Fourier transformation should be $$ X_f = |T| \frac{T}{1 + 2 i \pi f T} e^{- 2 i \pi f T } $$ 

On my book the result is the same except for T|T| but I don’t know what’s wrong. Thank you so much
If you already know the - quite well-known :) - FT pair $$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j2\pi f}$$ then by setting $a = 1/T$, you get $$e^{-\frac{t}{T}}u(t) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}$$
Time delay property $$x(t-t_0) \longleftrightarrow X(f)e^{-j2\pi ft_0}$$ can be used now by setting $t_0 = T$, and then $$x(t-T) = e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}e^{-j2\pi fT}$$
Adding your $1/T$ constant in front, you get $$\frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{\frac{1}{T}}{\frac{1}{T} + j2\pi f}e^{-j2\pi fT}$$
Finally, multiplying both numerator and denominator by $T$ you get what you want: $$\frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{1}{1 + j2\pi fT}e^{-j2\pi fT}$$