Fourier transform of the linear function

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Hello,
I was wondering if one can give meaning to the Fourier transform of the linear function:

[tex] \int_{-\infty}^{+\infty} x e^{ikx}\, dx [/tex]

I found that it is [tex] \frac{\delta(k)}{ik} [/tex], does this make sense?
 

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  • #2
vanhees71
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This expression doesn't make sense since it's intrinsically undefined. A handwaving way is
[tex]\int_{\mathbb{R}} \mathrm{d} x x \exp(\mathrm{i} k x)=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \int_{\mathbb{R}} \mathrm{d} x \exp(\mathrm{i} k x)=-2 \pi \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \delta(k).[/tex]
 
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  • #3
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This expression doesn't make sense since it's intrinsically undefined. A handwaving way is
[tex]\int_{\mathbb{R}} \mathrm{d} x x \exp(\mathrm{i} k x)=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \int_{\mathbb{R}} \mathrm{d} x \exp(\mathrm{i} k x)=-2 \pi \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \delta(k).[/tex]
Hmm.. seems to make sense. Why is there a minus sign popping up?
 
  • #4
mathman
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d/dk(exp(ikx)) = ixexp(ikx). you need -i to get 1 for the original integral.
 
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