# QM Fourier Transform and Integral rules

katkatkat
Show G(k)=$\sqrt{2π}$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)=$^{+∞}_{-∞}$∫dyf1(y)f2(x-y)

SO FAR

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

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

mult by e-ikyeiky

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

=g1(k)$^{+∞}_{-∞}$∫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)$^{+∞}_{-∞}$∫f2(x-y)e-ik(x-y)d(x-y)+g1(k)$^{+∞}_{-∞}$∫f2(x-y)e-ik(x-y)d(y)

=$\sqrt{2π}$g1(k)g2(k)+g1(k)$^{+∞}_{-∞}$∫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

=$\sqrt{2π}$g1(k)g2(k)

=G(k)

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## Answers and Replies

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

=g1(k)$^{+∞}_{-∞}$∫f2(x-y)e-ik(x-y)dx

This is not correct, you cannot claim that

$$\int_{ -\infty }^{ \infty }\int_{ -\infty }^{ \infty }dydx f_1(y)f_2(x-y) e^{-iky} e^{-ik(x-y)} = \left( \int_{ -\infty }^{ \infty }dx f_2(x-y) e^{-ik(x-y)} \right) \left( \int_{ -\infty }^{ \infty }dy f_1(y)e^{-iky} \right)$$

The first integral contains the variable $y$, so is not constant when integrating over $y$ and cannot be pulled out of the $y$ integral. Instead, try integrating over $x$ first while treating $y$ as a constant (which you can do since they are independent variables over the entire region of integration).

katkatkat
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)=$1/\sqrt{2π}$$^{+∞}_{-∞}$∫F(x)eikxdx

=$1/\sqrt{2π}$$^{+∞}_{-∞}$∫dx*eikx$^{+∞}_{-∞}$∫f1(y)f2(x-y)dy

=$1/\sqrt{2π}$$^{+∞}_{-∞}$∫d(x-y)f2(x-y)eik(x-y)$^{+∞}_{-∞}$∫f1(y)eikydy

=$\sqrt{2π}$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)

Homework Helper
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Just use a substitution: $u(x)=x-y$, when integrating over x, y is constant, so $du=?$...

Homework Helper
Gold Member
Show G(k)=$\sqrt{2π}$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)=$^{+∞}_{-∞}$∫dyf1(y)f2(x-y)

SO FAR

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

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

mult by e-ikyeiky

=$1/\sqrt{2π}$$^{+∞}_{-∞}$∫dyf1(y)e-iky$^{+∞}_{-∞}$∫f2(x-y)e-ik(x-y)dx
No, the problem starts here.

The second integral, which, when evaluated (integrated over x), being a function of y, therefore belongs under the 1st integral. You put the second integral outside the first.

For the last step, hint: ∫f2(x-y)exp{-ik(x-y)}dx = same expression with dx → d(x-y), since y is constant in the x integration.

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madhatter42
Hi,

I just wanted to note that you are essentially trying to prove the convolution theorem for the Fourier transform. Knowing that may help you out by looking at its proof.

katkatkat
The second integral, which, when evaluated (integrated over x), being a function of y, therefore belongs under the 1st integral. You put the second integral outside the first.

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.

katkatkat
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.

Staff Emeritus
Science Advisor
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You could use the change of variables u=y, v=x-y. It's straightforward to show that dx dy = du dv so that
$$\int_{-\infty}^\infty \int_{-\infty}^\infty dy\,dx\,f_1(y)\,f_2(x-y) e^{-iky} e^{-ik(x-y)} = \int_{-\infty}^\infty \int_{-\infty}^\infty du\,dv\,f_1(u)\,f_2(v) e^{-iku} e^{-ikv}.$$

Homework Helper
Gold Member
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.

Ahem - if it' a function of (x-y) it's ipso facto a function of x ... the second integral, when integrated wrt x, is now a function only of y.

You don't need to substitute anything, just realize that dx and d(x-y) are the same thing if y is a constant, which it is in the second integral (the one ending in dx). And you can't just say, "well, I'll get rid of y by substituting z = x-y." That's actually quite amusing ...