I'm trying to compute the Fourier transform of the function $u_y(x)=e^{-y|x|}$ and i've come up to calculating $I=\int_0^{\infty}e^{-(i\xi+y)x}dx$ for example. Can i say this integral is $\frac{e^{-(i\xi+y)x}}{i\xi+y}$ 'evaluated' from zero to $\infty$ ?
I thought this: considering the curve $\gamma:[0,\infty[\longrightarrow \mathbb{C}$ given by $\gamma(x)=(i\xi+y)x$ i have $I=\frac{1}{i\xi+y}\int_{\gamma}e^{-z}dz$ from the definition of integral along a parametrized path. So now i should say that $\int_{\gamma}e^{-z}dz=\int_0^{\infty}e^{-x}dz=1$. I should integrate over a closed path composed by $\gamma_1:[0,R] \longrightarrow \mathbb{C}$, $\gamma_1(x)=x$, $\gamma_2$ a part of a circle of radius R starting from $(R,0)$ up to the intersection with the curve $\gamma$. If i prove the integral over $\gamma_2$ tends to zero as R tends to $\infty$ i will have proved my assertion, but i can't do this. This is what i got: $\gamma_2(x)=Re^{ix}$ with domain $[0,T]$ for some T. Inserting into the integral (call it J) i come up with the estimate $J \leq R\int_0^T|e^{Re^{ix}}|dx =R\int_0^Te^{R\cos{x}}$ but it doesn't seem to approach zero as R goes to $\infty$!
(Anyway, if this is correct i get $\frac{2y}{y^2+\xi^2}$ as the fourier transform)