Discrete Fourier transform of an exponential decay
This boils down to computing the sum of a geometric series. Given

$$x[n]=e^{-\alpha n},\qquad n=0,1,\ldots,N-1\tag{1}$$

the DFT is

$$\begin{align}X[k]&amp;=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\\&amp;=\sum_{n=0}^{N-1}e^{-\alpha n}e^{-j2\pi nk/N}\\&amp;=\frac{1-e^{-\alpha N}}{1-e^{-\alpha}e^{-j2\pi k/N}},\qquad\alpha\neq 0,\quad k=0,1,\ldots N-1\tag{2}\end{align}$$

For $\alpha=0$, you have $X[0]=N$ and $X[k]=0$ for $k=1,2,\ldots,N-1$.

If you zero pad with $N$ zeros, i.e., if the DFT length is $2N$, the result can be computed in a very similar manner:

$$\begin{align}X[k]&amp;=\sum_{n=0}^{N-1}x[n]e^{-j\pi nk/N}\\&amp;=\sum_{n=0}^{N-1}e^{-\alpha n}e^{-j\pi nk/N}\\&amp;=\frac{1-e^{-jk\pi}e^{-\alpha N}}{1-e^{-\alpha}e^{-j\pi k/N}}\\&amp;=\frac{1-(-1)^ke^{-\alpha N}}{1-e^{-\alpha}e^{-j\pi k/N}},\qquad\alpha\neq 0,\quad k=0,1,\ldots 2N-1\tag{3}\end{align}$$

For $\alpha=0$ you get

$$X[k]=\begin{cases}N,\quad k=0\\0,\quad k\textrm{ even},k\neq 0\\\displaystyle\frac{2}{1-e^{-j\pi k/N}},
\quad k\textrm{ odd}\end{cases}\tag{4}$$