Considering the function $g(x) = e^{-x}, x>0$ and $g(-x) = -g(x)$. I am trying to find the Fourier integral representation of $g(x)$.
I am looking for some clarification on the progress I have made below to see if I am on the correct path before I begin integrating.
So from the question, $g(-x) = -g(x)$ meaning $g(x)$ is an odd function. Then,
$$F(\omega)=\int_{-\infty}^{\infty}f(x)e^{-j\omega x}dx$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega x}d\omega$$
Expand the first formula yielding
$F(\omega)=A(\omega)+jB(\omega)$
where, $A(\omega)=\int_{-\infty}^{\infty}f(x)\cos(\omega x)dx$, $B(\omega)=-\int_{-\infty}^{\infty}f(x)\sin(\omega x)dx$
Since $g(x)$ is an odd function, $A(\omega)=0$. Then now what I have to do is calculate $B(\omega)$
$$B(\omega)=-\int_{-\infty}^{\infty}f(x)\sin(\omega x)dx=-2\int_{0}^{\infty}e^{-x}\sin(\omega x)dx$$
So I am wondering if I have taken the correct steps, thanks!
I see from the comments that the wording may have confused you. $g(x)$ is defined to be the odd extension of $e^{-x}$ on the entire real line. This means its Fourier representation is a sine integral
$$ g(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} B(\omega) \sin (\omega x) \ d\omega $$
where
$$ B(\omega) = \int_{-\infty}^{\infty} g(x) \sin(\omega x) \ dx = -\int_{-\infty}^0 e^x \sin (\omega x) + \int_0^{\infty} e^{-x} \sin (\omega x) \ dx \\ = 2\int_0^{\infty} e^{-x} \sin (\omega x) \ dx$$