Fourier transform of the linear function

[Irid]

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

$$\int_{-\infty}^{+\infty} x e^{ikx}\, dx$$

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

 

[vanhees71]

This expression doesn't make sense since it's intrinsically undefined. A handwaving way is
$$\int_{\mathbb{R}} \mathrm{d} x x \exp(\mathrm{i} k x)=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \int_{\mathbb{R}} \mathrm{d} x \exp(\mathrm{i} k x)=-2 \pi \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \delta(k).$$

 

[Irid]

This expression doesn't make sense since it's intrinsically undefined. A handwaving way is
$$\int_{\mathbb{R}} \mathrm{d} x x \exp(\mathrm{i} k x)=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \int_{\mathbb{R}} \mathrm{d} x \exp(\mathrm{i} k x)=-2 \pi \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} k} \delta(k).$$

Hmm.. seems to make sense. Why is there a minus sign popping up?

 

[mathman]

d/dk(exp(ikx)) = ixexp(ikx). you need -i to get 1 for the original integral.