Help in proving some Fourier transform property

Question

Knowing the integral property: $$\int_{-\infty}^tx(\tau)\,d\tau\,\overset{F}\longleftrightarrow\,\frac{X(\omega)}{j\omega}\,+\,\pi X(0)\delta(\omega)$$ and the inversion theorem: $$f(t)\,\overset{F}\longleftrightarrow\,g(\omega)\,\Rightarrow\,g(t)\,\overset{F}\longleftrightarrow\,2\pi f(-\omega)$$

How can we prove that: $$\frac{-x(t)}{jt}\,+\,\pi x(0) \delta(t)\,\overset{F}\longleftrightarrow\,\int_{-\infty}^\omega X(\eta)\,d\eta$$

Here the $\,\overset{F}\longleftrightarrow\,$ denotes the Fourier transform defined as: $$X(\omega)\,=\,\int_{-\infty}^\infty\,x(t)\,e^{-j\omega t}\,dt$$So:$$\Rightarrow\,x(t)\,=\,\frac{1}{2\pi}\,\int_{-\infty}^\infty\,X(\omega)\,e^{j\omega t}\,d\omega$$ My question stems from the signals and systems course in the electrical engineering field, so here the Fourier transform should give us all of the available frequencies in the signal $x(t)$ ($x(t)$ models the signal's behavior in the time dimension): $$X(\omega)=2\pi.(\frac{available\,frequency}{d\omega})=2\pi.(frequency\,density)$$ Also the $\delta(x)$ is the Dirac delta function.
Here are two other properties that might be helpful in solving this question: $$x(-t)\,\overset{F}\longleftrightarrow\,X(-\omega)$$ $$x^*(t)\,\overset{F}\longleftrightarrow\,X^*(-\omega)$$ By $x^*(t)$ I mean the complex conjugate of $x(t)$.
Let me know if anything needs further explanations.

Answer 1

I assume that $f(t)\,\overset{F}\longleftrightarrow\,g(\omega)$ means that the Fourier transform of $f(t)$ is $g(\omega)$: $$ g(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt $$ By the Fourier inversion theorem one has $$ f(t)\,\overset{F}\longleftrightarrow\,g(\omega)\,\Rightarrow\,g(t)\,\overset{F}\longleftrightarrow\,2\pi\,f(-\omega) $$ so since $$ \int_{-\infty}^tx(\tau)\,d\tau \,\overset{F}\longleftrightarrow\, \frac{X(\omega)}{j\omega}+\pi X(0)\delta(\omega) $$ where $x(t)\,\overset{F}\longleftrightarrow\,X(\omega)$ you get $$ \frac{X(t)}{jt}+\pi X(0)\delta(t) \,\overset{F}\longleftrightarrow\, 2\pi\int_{-\infty}^{-\omega} x(\tau)\,d\tau $$ Now we want to replace $X(t)$ with $x(t)$, but we need a Fourier transform in the left hand side. Luckily, since $x(t)\,\overset{F}\longleftrightarrow\,X(\omega)$, by the Fourier inversion theorem we have $X(t)\,\overset{F}\longleftrightarrow\,2\pi\,x(-\omega)$ so by replacing $X(t)$ with $2\pi\,x(-t)$ in the left hand side we can replace $x(\tau)$ with $X(\tau)$ in the right hand side: $$ \frac{2\pi\,x(-t)}{jt}+\pi\,2\pi x(-0)\delta(t) \,\overset{F}\longleftrightarrow\, 2\pi\int_{-\infty}^{-\omega} X(\tau)\,d\tau $$ or after division with $2\pi$: $$ \frac{x(-t)}{jt}+\pi\,x(-0)\delta(t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{-\omega} X(\tau)\,d\tau $$ Finally, changing $t$ to $-t$ in the left hand side is equivalent to changing $\omega$ to $-\omega$ in the right hand side, so we get $$ \frac{x(t)}{j(-t)}+\pi\,x(0)\delta(-t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{\omega} X(\tau)\,d\tau $$ i.e. $$ -\frac{x(t)}{jt}+\pi\,x(0)\delta(t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{\omega} X(\tau)\,d\tau $$