Use integration by parts to show that

\[
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\omega \sin (t \omega) d \omega}{\left(1+\omega^{2}ight)^{v+3 / 2}} & =\frac{t}{2 v+1} \int_{-\infty}^{\infty} \frac{\cos (t \omega) d \omega}{\left(1+\omega^{2}ight)^{v+1 / 2}} \\
& =\frac{\sqrt{ } \pi}{2^{v-1} \Gamma(v+1 / 2)} \times \frac{t}{2 v+2}|t|^{v} \mathcal{K}_{v}(t)
\end{aligned}
\]

Hence deduce that, for any pair of linear functionals \(v, x: \mathbb{R}^{d} ightarrow \mathbb{R}\),

\[
\int_{\mathbb{R}^{d}} \frac{\omega v \sin (\omega x) d \omega}{\left(1+\|\omega\|^{2}ight)^{v+d / 2+1}}=\frac{\pi^{d / 2}}{2^{v-1} \Gamma(u+d / 2)} \times \frac{v x}{2 v+1}\|x\|^{v} \mathcal{K}_{v}(\|x\|)
\]

where \(v x\) denotes the scalar product.