I was reading a paper and I came across one equation, in which I had a problem deriving this equation.
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{G(\omega-\omega_c)}{j\omega}e^{j\omega(\frac{p-x_0}{s})}\,d\omega=\int_{-\infty}^{\frac{p-x_0}{s}}g(x)e^{j\omega_cx}\,dx$$, where $G(\omega)$ is the fourier transform of $g(x)$.
Why is this so?
Let $f^{\wedge}$ be the Fourier transform and $g^{\vee}$ be the inverse Fourier transform. Let $f\star g$ be the convolution. Let $g$ be your $g$ and let $f=\chi_{a,b}$ be the function which is $1$ on $[a,b]$ and $0$ elsewhere. The integral on the right is related to $(fg)^{\vee}$. But $\sqrt{2\pi}fg=(f^{\vee}\star g^{\vee})^{\wedge}$ gives $$ \sqrt{2\pi}(fg)^{\vee}=f^{\vee}\star g^{\vee}. $$ And $$ f^{\vee}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{a}^{b}e^{j\omega x}dx=\frac{1}{\sqrt{2\pi}j\omega}(e^{j\omega b}-e^{j\omega a}) $$ Let $G(\omega)=g^{\wedge}(\omega)$ as you have stated. Then $G(-\omega)=g^{\vee}(\omega)$ gives you $$ \int_{a}^{b}g(x)e^{j\omega_{c} x}\,dx=\sqrt{2\pi}(fg)^{\vee}(\omega_{c})=(f^{\vee}\star g^{\vee})(\omega_{c})=\int_{-\infty}^{\infty} \frac{e^{jb\omega}-e^{ja\omega}}{j\omega}G(\omega-\omega_{c})d\omega. $$ Now let $a\rightarrow-\infty$ to obtain $$ \int_{-\infty}^{b}g(x)e^{j\omega_{c}x}\,dx = \int_{-\infty}^{\infty}\frac{G(\omega-\omega_{c})}{j\omega}e^{j\omega b}\,d\omega. $$ To take that last limit, you'll need some conditions on $G(\omega-\omega_{c})/\omega$ such as integrability. But that's the idea. Let $b=(p-x_{0})/s$ to finish up. I have the factors of $2\pi$ messed up, but I'll leave that to you to straight out. I'm always getting those things reversed in convolution.
