Fourier transform of $e^{-t}u(t)$

Question

I have this signal $ x(t) = \frac{1}{T} e^{- \frac{-(t-T)}{T } } u(t - T ) $ and I found that the Fourier transform is $ X(f) = \frac{ e^{-i \pi 2 f T} e }{ 1 + i 2 \pi f T } $. The correct result is without the $ e^1 $ but I didn't know why , I made a decomposition before to apply the Fourier transform

Answer 1

$\require{cancel}$

\begin{eqnarray} X(f) &=& \int_{-\infty}^{+\infty}{\rm d}t~ \frac{1}{T}e^{-(t-T)/T}u(t- T) e^{-2\pi i f t} \\ &=& \frac{1}{T} \int_{T}^{+\infty}{\rm d}t~ e^{T/T} e^{-(1 + 2\pi i f T)t/T} \\ &=& \frac{e}{\cancel{T}}\frac{\cancel{T}}{1 + 2\pi i f T} e^{-(1 + 2\pi i f T)} \\ &=& \frac{\cancel{e}}{1 + 2\pi i f T} \cancel{e^{-1}} e^{-2\pi i f T} \\ &=& \frac{e^{-2\pi if T}}{1 + 2\pi i f T} \end{eqnarray}