Let us consider following example
$$x(at)\iff\dfrac1{|a|}X\left(j\left(\dfrac\omega a\right)\right).$$
one thing which i did not understand is where absolute value of $a$ came from? Ok if we have
$$\int^{\infty}_{-\infty} x(a*t)*e^{-j\omega*t}dt$$
then we may have denote $y=a*t$ from which $t=\frac {y} {a}$ and also $dy=dt*a$ or $$dt=\frac {dy} {a}$$ so when we will put all this thing we will get
$$\frac{1}{a}\int^{\infty}_{-\infty} x(y)*e^{-j\omega*\frac {y}{a}}dy$$ this should equal
$$\frac{1}{a}*X\left(j*\left(\frac{\omega}{a}\right)\right)$$ what about absolute value?thanks in advance
You have supposed that $a>0$.
You have to take cases for $a$.
In the case of $a<0$ the absolute value will appear.
$a<0:$
$$\int_{-\infty}^{+\infty}x(a \cdot t) \cdot e^{-j \omega t}dt$$ $$y=a \cdot t \Rightarrow t=\frac{y}{a}$$ $$dy=adt$$ $$\text{ When } t \rightarrow -\infty \Rightarrow y \rightarrow +\infty$$ $$\text{ When } t \rightarrow +\infty \Rightarrow y \rightarrow -\infty$$
$$\int_{-\infty}^{+\infty}x(a \cdot t) \cdot e^{-j \omega t}dt=\frac{1}{a}\int_{+\infty}^{-\infty} x(y) \cdot e^{-j \omega \frac{y}{a}} dy=-\frac{1}{a}\int_{-\infty}^{+\infty} x(y) \cdot e^{-j \frac{\omega}{a} y} dy=\frac{1}{-a}X(j(\frac{\omega}{a}))=\frac{1}{|a|}X(j(\frac{\omega}{a}))$$
Since $a<0$, $-a$ is the positive value of $a$, so it equal to $|a|$.