Calculation of Fourier transform How to calculate the Fourier transform of $f(x)=x$.
I know using the formula $f(\varepsilon)=\int_xe^{-ix\varepsilon}x \, dx$.
But I have problem calculating this complex integral. 
 A: Hint: $\frac{d}{dx} e^{-i\epsilon x} = -i\epsilon e^{-i\epsilon x}$ and use integration by parts.
A: As said before, $f$ is not in $L^1$, so we cannot calculate that integral directly.
The way out is to use a test function $g \in C^\infty_c$ (compactly supported and smooth). Further, i assume you are familar with Plancherel's theorem.
Then, we can calculate
$$\langle \mathcal{F}x, g\rangle = \iint e^{ix\xi}xg(\xi)dxd\xi\\=i\iint e^{-ix\xi}g'(\xi)dxd\xi = \langle 1, \mathcal{F}(i\partial_\xi g)\rangle$$ using integration by parts.
Now the question remains: What is $\langle 1, \mathcal{F}f\rangle$?
To answer this define the usual delta-distribution: $\delta : f \mapsto f(0)$ for $f \in C^\infty_c$.
Then calculate (using Plancherel):
$$\langle \mathcal F \delta, f \rangle = \langle \delta, \mathcal{F} f\rangle=\mathcal{F}f(0) = \langle 1, f\rangle.$$
Then it follows (using Plancherel again) that $\langle\mathcal{F}x,g\rangle = i\partial_\xi g(0)$.
So in that sense we could write that $\mathcal{F}x = -i\partial_\xi \delta$ (by integration by parts).
A: Use integration by parts to get
$$\int{e^{-ix\epsilon}xdx}=-\frac{1}{i\epsilon}e^{-ix\epsilon}x+\frac{1}{i\epsilon}\int{e^{-ix\epsilon}dx}=\frac{i}{\epsilon}e^{-ix\epsilon}x-\frac{i}{\epsilon}\int{e^{-ix\epsilon}dx}$$
A: Ss others have pointed out, $f(x)$ is not in $L^1(\mathbb{R})$ so the function would need to be restricted to some compact interval and/or made periodic and then extended onto $\mathbb{R}$
