Fourier transform of a tempered distribution ($\hat{u}$)

With $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, the Fourier transform of a tempered distribution is defined byContinuity of the Fourier transform on the Schwartz space makes this well defined. For regular integrable functions, Fubini's theorem shows that it agrees with the ordinary transform. The inverse convention has factor $(2\pi {)}^{-n}$ and the positive exponential.

For $D_j=-i{\partial }_j$, the Fourier transform of a tempered distribution satisfiesThe base cases follow from $⟨D_{j}u,\phi ⟩=i⟨u,{\partial }_j\phi ⟩$ and integration by parts in the Fourier integral. Iterating gives the multi-index formulas. The $D$ convention matters: for the unscaled derivative, $\widehat{{\partial }_{j}u}=i{\xi }_j\hat{u}$.

For $u(x)=\frac{1}{2}\log (1+x^2)$, the angular-frequency transform is the tempered distributionThe numerator cancels the singularity at zero and makes this integral absolutely convergent. Differentiating $u$ and using $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\xi |}$ determines this expression up to a Dirac delta distribution. That delta coefficient is zero: testing with the expanding Gaussian ${\phi }_n(x)=(2\sqrt{\pi }{)}^{-1}{e}^{-x^2/(4n)}$ makes both the proposed expression and $⟨u,{\hat{\phi }}_n⟩$ tend to zero, while ${\phi }_n(0)$ stays fixed. Changing the subtraction convention would change the delta coefficient.