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!