Homework Help: QM Fourier Transform and Integral rules

Show G(k)=[itex]\sqrt{2π}[/itex]g₁(k)g₂(k)

Given that G(k) is the fourier transform of F(x), g₁(k) is fourier trans of f₁(x), g₂(k) is fourier trans of f₂(X) and

F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf₁(y)f₂(x-y)

SO FAR

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e^-ikxdx <-def'n of fourier transform

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf₁(y)f₂(x-y)e^-ikxdx

mult by e^-ikye^iky

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf₁(y)e^-iky[itex]^{+∞}_{-∞}[/itex]∫f₂(x-y)e^-ik(x-y)dx

=g₁(k)[itex]^{+∞}_{-∞}[/itex]∫f₂(x-y)e^-ik(x-y)dx

Problem starts here... I have arg of x-y in the remaining part, with dx...
Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this

=g₁(k)[itex]^{+∞}_{-∞}[/itex]∫f₂(x-y)e^-ik(x-y)d(x-y)+g₁(k)[itex]^{+∞}_{-∞}[/itex]∫f₂(x-y)e^-ik(x-y)d(y)

=[itex]\sqrt{2π}[/itex]g₁(k)g₂(k)+g₁(k)[itex]^{+∞}_{-∞}[/itex]∫f₂(x-y)e^-ik(x-y)d(y)

I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result

=[itex]\sqrt{2π}[/itex]g₁(k)g₂(k)

=G(k)

 

I can't actually solve the integral as I'm supposed to be subbing in the symbol for the fourier transform. I have no choice but to separate them somehow.

Here's the prof's solution. I'm not sure how he got to the 3rd step though. How did he get d(x-y) out of dx.

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e^ikxdx

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dx*e^ikx[itex]^{+∞}_{-∞}[/itex]∫f₁(y)f₂(x-y)dy

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫d(x-y)f₂(x-y)e^ik(x-y)[itex]^{+∞}_{-∞}[/itex]∫f₁(y)e^ikydy

=[itex]\sqrt{2π}[/itex]g₁(k)g₂(k)

I know his solution is right, this is coming from someone who's specialty is fourier transforms. I'm just not sure how he did the d(x-y) step from the dx step (line 2 to line 3)

 

I don't think that matters. It's a function of the variable (x-y) not y. Can you not sub in z=x-y? Then you no longer have that problem. I don't think it'll fall under the dy integral for that reason since it's not a function of the y variable.

 

I'm going to sub in z(x)=x-y. Then dx=dz and f(x-y)=f(z) etc. Thanks for the help guys :) This looks on the right track according to some of the proofs I've seen.