Fourier integration of $f(x)=\pi e^{-x}$

Question

I've tried below Fourier integration and reached some answer.I would appreciate if anyone takes a look at this and enlighten me if something is wrong (or if it is right):

$ f(x)=\begin{cases}\pi e^{-x} & x>0 \\ 0 & x<0 \end{cases} $

My answers are :

$ A(w)=\dfrac{1}{1+w^{2}} $

$ B(w)=\dfrac{w}{1+w^{2}} $

finally :

$ f(x)=\int_0^\infty \dfrac{1}{1+w^{2}} cos(wx)+\dfrac{w}{1+w^{2}}sin(wx)dx $

Answer 1

Hint. You may observe that $$ \int_0^{+\infty}e^{-a \:x}dx=\frac1a,\qquad \Re a>0. \tag1 $$ Then, for any $w \in \mathbb{R}$, you have $$ \int_{-\infty}^{+\infty}f(x)\:e^{-i\:w \:x}\:dx=\int_0^{+\infty}\pi e^{-a \:x}e^{-i\:w \:x}\:dx= \pi\:\frac1{1+w^2}-i \pi\:\frac{ w}{1+w^2}. $$