Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

A) $\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$
B) $\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$
C) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$
D) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$
E) none of the above