The function is, $f(x) = \begin{cases}0 & x < 0\\e^{-x} & x > 0\end{cases}$
I have to find the fourier integral representation and hence show that
$$\int_{0}^{\infty} \frac{\cos\omega{x}+\omega{\sin\omega{x}}}{1+\omega^{2}}dw
= \begin{cases}0 & x < 0\\\frac{\pi}{2} & x = 0\\\pi{e^{-x}} & x>0\end{cases}$$
Edit:
The fourier integral representation of a function is defined as follows:
$$ f(x)=\int_{0}^{\infty} [A(w)coswx+B(w)sinwx] dw $$
where
$$A(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)coswv]dv$$
$$B(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)sinwv]dv$$
Best Answer
Take the fourier transform using the definition as an integral,
$$\mathcal{F} (f(x)) = \int_{\mathbb{R}} f(x)e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x}e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x(2\pi i\xi + 1)}dx$$
Now just evaluate the integral to get,
$$\mathcal{F} (f(x)) = \frac{e^{-x(2\pi i \xi +1)}}{-(2\pi i \xi+1)}\biggr|_0^\infty = \frac{1}{(2\pi i \xi +1)}$$