# How to find the fourier integral of this function?

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

$$A(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)coswv]dv$$ $$B(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)sinwv]dv$$

• By fourier integral do you mean fourier transform or something else? – TSF Feb 5 '18 at 12:31
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Feb 5 '18 at 12:33

$$\mathcal{F} (f(x)) = \int_{\mathbb{R}} f(x)e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x}e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x(2\pi i\xi + 1)}dx$$
$$\mathcal{F} (f(x)) = \frac{e^{-x(2\pi i \xi +1)}}{-(2\pi i \xi+1)}\biggr|_0^\infty = \frac{1}{(2\pi i \xi +1)}$$