Show G(k)=[itex]\sqrt{2π}[/itex]g(adsbygoogle = window.adsbygoogle || []).push({}); _{1}(k)g_{2}(k)

Given that G(k) is the fourier transform of F(x), g_{1}(k) is fourier trans of f_{1}(x), g_{2}(k) is fourier trans of f_{2}(X) and

F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf_{1}(y)f_{2}(x-y)

SO FAR

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

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

mult by e^{-iky}e^{iky}

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

=g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(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

=g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}d(x-y)+g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(x-y)e^{-ik(x-y)}d(y)

=[itex]\sqrt{2π}[/itex]g_{1}(k)g_{2}(k)+g_{1}(k)[itex]^{+∞}_{-∞}[/itex]∫f_{2}(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]g_{1}(k)g_{2}(k)

=G(k)

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# Homework Help: QM Fourier Transform and Integral rules

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