Let $$f(t) = \begin{cases}e^{-\frac{|t|}{T}}&if&|t| \lt T \\0&if &|t| \ge T\end{cases}$$
I believe I have to separate this into three integrals, from $-\infty$ to -T, from -T to T, and from T to $\infty$. But I get confused with the absolute value, and how to separate that.
Let us define the unitary Fourier transform of $f$: \begin{align} \hat{f}\left(\omega\right)&:=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f\left(t\right)\mathrm{e}^{-\mathrm{i}\omega t}\mathrm{d}t. \end{align} We compute: \begin{align} \hat{f}\left(\omega\right)&=\frac{1}{\sqrt{2\pi}}\int_{-T}^{T}\mathrm{e}^{-\frac{\left|t\right|}{T}}\mathrm{e}^{-\mathrm{i}\omega t}\mathrm{d}t\\ &=\frac{1}{\sqrt{2\pi}}\int_{-T}^{0}\mathrm{e}^{\frac{t}{T}}\mathrm{e}^{-\mathrm{i}\omega t}\mathrm{d}t+\frac{1}{\sqrt{2\pi}}\int_{0}^{T}\mathrm{e}^{-\frac{t}{T}}\mathrm{e}^{-\mathrm{i}\omega t}\mathrm{d}t\\ &=\frac{1}{\sqrt{2\pi}}\int_{-T}^{0}\mathrm{e}^{\left(\frac{1}{T}-\mathrm{i}\omega\right)t}\mathrm{d}t+\frac{1}{\sqrt{2\pi}}\int_{0}^{T}\mathrm{e}^{-\left(\frac{1}{T}+\mathrm{i}\omega\right)t}\mathrm{d}t\\ &=\frac{1}{\sqrt{2\pi}}\left[\frac{\mathrm{e}^{\left(\frac{1}{T}-\mathrm{i}\omega\right)t}}{\frac{1}{T}-\mathrm{i}\omega}\right]_{t=-T}^{t=0}-\frac{1}{\sqrt{2\pi}}\left[\frac{\mathrm{e}^{-\left(\frac{1}{T}+\mathrm{i}\omega\right)t}}{\frac{1}{T}+\mathrm{i}\omega}\right]_{t=0}^{t=T}\\ &=\frac{1}{\sqrt{2\pi}}\left[\frac{2\mathrm{i}\mathrm{e}^{-\left(\frac{1}{T}-\mathrm{i}\omega\right)\frac{T}{2}}}{\frac{1}{T}-\mathrm{i}\omega}\sin\left(\left(\frac{1}{T}-\mathrm{i}\omega\right)\frac{T}{2}\right)-\frac{2\mathrm{i}\mathrm{e}^{-\left(\frac{1}{T}+\mathrm{i}\omega\right)\frac{T}{2}}}{\frac{1}{T}+\mathrm{i}\omega}\sin\left(\left(\frac{1}{T}+\mathrm{i}\omega\right)\frac{T}{2}\right)\right]. \end{align}