Having trouble finding the fourier transform of the function $f_a(x)=e^{-a|x|}$ , where $a>0$
I currently have that $$ \mathcal{F}(f_ax) = \int_{0}^{\infty} e^{-ax}\, e^{ i s x} \,ds= \dfrac{-1}{is-a}\sqrt{\dfrac2{\pi}} $$
I have missed out the working since it takes me too long to format everything, however this is the answer I got, however I dont believe I'm right, any help would be greatly appreciated.
You missed half of the fourier integral. We have$\def\F{\mathcal F}$ \begin{align*} \F(f_a)(s) &= \frac 1{(2\pi)^{1/2}}\int_{-\infty}^\infty \exp\bigl(-a|x|\bigr)\exp(-isx)\, dx\\ &= \frac 1{(2\pi)^{1/2}} \int_0^\infty \exp\bigl(-(is+a)x\bigr)\, dx + \frac 1{(2\pi)^{1/2}} \int_{-\infty}^0 \exp\bigl(-(is-a)x\bigr)\, dx\\ &= \frac 1{(2\pi)^{1/2}} \frac 1{a + is} + \frac 1{(2\pi)^{1/2}} \frac 1{a-is}\\ &= \frac 1{(2\pi)^{1/2}} \frac{2a}{a^2 + s^2}\\ &= \sqrt{\frac 2\pi} \frac a{a^2 + s^2} \end{align*}