Fourier Transform of integral of a signal

[maverick280857]

Hi. I have a question regarding the continuous time Fourier Transform of an input signal:

$$x(t) \rightarrow X(j\omega)$$

then

$$\int_{-\infty}^{t}x(\tau)d\tau \rightarrow \frac{X(j\omega)}{j\omega} + \pi X(0)\delta(\omega)$$

but if I want to write it in terms of $f = \frac{\omega}{2\pi}$, should it be:

$$\int_{-\infty}^{t}x(\tau)d\tau \rightarrow \frac{X(j\omega)}{j\omega} + \frac{1}{2}X(0)\delta(f)$$

How does the $\pi$ get replaced by $\frac{1}{2}$ here?

 

[George Jones]

How does the $\pi$ get replaced by $\frac{1}{2}$ here?[/QUOTE]

For $a>0$, what does $\delta (ax )$ equal? Why?

 

[maverick280857]

Oh ok, so

$\delta(\omega) = \delta(2\pi f) = \frac{1}{2\pi}\delta(f)$

Thanks