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 a $2\pi$ periodic distribution $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$, choose a smooth periodic partition of unity $\Psi$, and define $c_n=(2\pi {)}^{-1}⟨u,\Psi e^{-inx}⟩$. A finite Schwartz seminorm bound on the fixed compact support gives $|c_n|\le C(1+|n|{)}^N$. For a Schwartz function $\varphi$, the Poisson summation formula and Fourier inversion imply $P\hat{\varphi }(x)={\sum }_n\varphi (-n)e^{inx}$. Localizing the pairing for $u$ givesproving the displayed Fourier transform formula. The series converges in tempered distributions because its coefficients have polynomial growth while $\varphi (n)$ decreases rapidly. Applying the inverse Fourier transform gives $u={\sum }_n{c}_n{e}^{inx}$ in tempered distributions. These are the smooth-window Fourier coefficients of a periodic distribution, and the transform formula shows independence of $\Psi$.

The distributional derivative of $\log |x|$ is the principal-value reciprocal distribution. Its transform is $-i\pi \mathrm{sgn}k$. Dividing the derivative-transform relation by $ik$ gives the displayed finite-part inverse-absolute-value distribution, with an undetermined delta term fixed by the finite-part convention.

With the $e^{-ikx}$ convention, exponential damping on the positive half-line gives $1/(a+ik)$. Its even part tends to $\pi \delta (k)$ and its odd part to $-i\mathrm{PV}(1/k)$. Thus the Heaviside step function transform requires a Cauchy principal value, not an ordinary reciprocal function at zero.

A real simple pole is not locally integrable in the ordinary improper sense. Its symmetric Cauchy principal value defines a tempered distribution. For the $e^{-ikx}$ transform convention, contour indentation or sine integration gives the displayed nondecaying oscillatory contribution. Upper and lower boundary-value prescriptions instead change it by $\mp i\pi \delta (x-a)$ before transforming. A pole prescription must therefore be stated before claiming a Fourier asymptotic expansion.

For an orthogonal matrix $R$, a unit-Jacobian change of variables gives $\widehat{\phi \circ R^t}=\hat{\phi }\circ R^t$. The dual definition of the Fourier transform of a tempered distribution consequently givesSince the Fourier transform is invertible on the Schwartz space and its dual, a tempered distribution is invariant under a rotation group exactly when its transform is invariant. This includes radial tempered distributions.

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.