The <Schwartz space> consists of the <smooth functions> $\varphi$ for which
$$
p_{\alpha,\beta}(\varphi)=\sup_{x\in\mathbb R^n}|x^\alpha\partial^\beta\varphi(x)|<\infty
$$
for all <multi-indices> $\alpha,\beta$. Its topology is generated by these <seminorms>; $\varphi_j\to\varphi$ means convergence in every seminorm. A <tempered distribution> is a continuous linear functional on this space. Convergence in $\mathcal S'$ here means weak convergence: $\langle u_j,\varphi\rangle\to\langle u,\varphi\rangle$ for every Schwartz function.

Use the unnormalized angular-frequency <Fourier transform>
$$
\widehat\varphi(\lambda)=\int_{\mathbb R^n}e^{-ix\cdot\lambda}\varphi(x)\,dx,
\qquad \langle\widehat u,\varphi\rangle=\langle u,\widehat\varphi\rangle.
$$
The transform is a continuous automorphism of the <Schwartz space>, so this duality defines a <tempered distribution>. Put $D_j=-i\partial_j$, as required by the stated identities. <Integration by parts> and differentiation under the integral give
$$
D_x^\alpha\widehat\varphi(x)=(-1)^{|\alpha|}\widehat{\lambda^\alpha\varphi}(x),
\qquad \widehat{D_\lambda^\beta\varphi}(x)=x^\beta\widehat\varphi(x).
$$
Thus, keeping the transpose factor of $D=-i\partial$ explicit,
$$
\langle\widehat{D_j u},\varphi\rangle
=i\langle u,\partial_{x_j}\widehat\varphi\rangle
=\langle\widehat u,\lambda_j\varphi\rangle,
$$
and
$$
\langle\widehat{x_j u},\varphi\rangle
=\langle u,x_j\widehat\varphi\rangle
=\langle\widehat u,D_j\varphi\rangle
=\langle-D_j\widehat u,\varphi\rangle.
$$
Iterating gives the <Fourier derivative identities for distributions>
$$
\boxed{\widehat{D^\alpha u}=\lambda^\alpha\widehat u,
\qquad \widehat{x^\beta u}=(-1)^{|\beta|}D^\beta\widehat u.}
$$

The logarithm defines a <tempered distribution> because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u'(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2)^{-1}}=\pi e^{-|\lambda|}$, obtainable by the <residue theorem>, gives
$$
\widehat{u'}(\lambda)=i\partial_\lambda(\pi e^{-|\lambda|})
=-i\pi\operatorname{sgn}(\lambda)e^{-|\lambda|}.
$$
As $\widehat{u'}=i\lambda\widehat u$, we have $\lambda\widehat u=-\pi\operatorname{sgn}(\lambda)e^{-|\lambda|}$. Define
$$
\langle T,\varphi\rangle
=-\pi\int_{\mathbb R}\frac{\varphi(\lambda)-\varphi(0)}{|\lambda|}e^{-|\lambda|}\,d\lambda.
$$
This integral is absolutely convergent: the numerator is $O(|\lambda|)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The <kernel of multiplication by a coordinate> therefore gives $\widehat u=T+c\delta_0$.

To determine $c$, use the supplied <Gaussian Fourier transform> pair with $\varphi_n(0)=1/(2\sqrt\pi)$. Then
$$
\langle\widehat u,\varphi_n\rangle
=\sqrt n\int_{\mathbb R}\tfrac12\log(1+x^2)e^{-nx^2}\,dx
=\int_{\mathbb R}\tfrac12\log(1+y^2/n)e^{-y^2}\,dy\longrightarrow0.
$$
Also $\langle T,\varphi_n\rangle\to0$ by the <dominated convergence theorem>: the relevant integrand is bounded by a constant times $|\lambda|e^{-|\lambda|}$ for $n\geq1$. Consequently
$$
\boxed{c=0,\qquad
\langle\widehat u,\varphi\rangle
=-\pi\lim_{\varepsilon\downarrow0}\int_{|\lambda|>\varepsilon}
\frac{\varphi(\lambda)-\varphi(0)}{|\lambda|}e^{-|\lambda|}\,d\lambda.}
$$
The cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is <Fourier transform of the logarithm of one plus x squared>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 327 / 1 / Solution

{c}
{title2=$\mathcal F[\tfrac12\log(1+x^2)]$}

For $u(x)=\tfrac12\log(1+x^2)$, the angular-frequency transform is the <tempered distribution>
$$
\langle\widehat u,\varphi\rangle
=-\pi\int_{\mathbb R}\frac{\varphi(\xi)-\varphi(0)}{|\xi|}e^{-|\xi|}\,d\xi.
$$
The 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 $\varphi_n(x)=(2\sqrt\pi)^{-1}e^{-x^2/(4n)}$ makes both the proposed expression and $\langle u,\widehat\varphi_n\rangle$ tend to zero, while $\varphi_n(0)$ stays fixed. Changing the subtraction convention would change the delta coefficient.

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 Incoming links: Fourier transform of the logarithm of one plus x squared