Fourier transform question, keep getting zero, minus infinity limit

[rwooduk]

calculate the Fourier transform of the function g(x) if g(x) = 0 for x<0 and g(x) = $e^{-x}$ otherwise.

putting g(x) into the transform we have:

$\tilde{g}(p) \propto \int_{0}^{inf} e^{-ipx} e^{-x} dx$

which we can write:

$\tilde{g}(p) \propto \int_{0}^{inf} e^{-x(ip+1)} dx$

which will give:

$\tilde{g}(p) \propto e^{-x(ip+1)}$ for x between 0 and infinity

the problem is $e^{0} = 1$ and $e^{-inf} = 1$ so i get zero.

is there a way around this?

thanks for any help.

 

[SteamKing]

calculate the Fourier transform of the function g(x) if g(x) = 0 for x<0 and g(x) = $e^{-x}$ otherwise.

putting g(x) into the transform we have:

$\tilde{g}(p) \propto \int_{0}^{inf} e^{-ipx} e^{-x} dx$

which we can write:

$\tilde{g}(p) \propto \int_{0}^{inf} e^{-x(ip+1)} dx$

which will give:

$\tilde{g}(p) \propto e^{-x(ip+1)}$ for x between 0 and infinity

the problem is $e^{0} = 1$ and $e^{-inf} = 1$ so i get zero.

is there a way around this?

thanks for any help.

Is $e^{-∞} = 1$? That would mean $1/e^{∞} = 1$, which implies $e^{∞} = 1$.

 

[rwooduk]

Is $e^{-∞} = 1$? That would mean $1/e^{∞} = 1$, which implies $e^{∞} = 1$.

oh dear, so easy, and it took me ages to do all the latex on that op.

Thanks for clearing this up it's appreciated!