We consider a function $f$ in $\mathcal{B}_1$ whose partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. We set: $$\forall q \in \mathbb{R}^+,\quad F(q) = \frac{1}{2\pi}\int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta = 2\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$$ We admit that we can interchange the two integrals and therefore that $$\forall \varepsilon > 0 \quad \int_\varepsilon^{+\infty} \left(\frac{1}{q^2}\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r\right)\mathrm{d}q = \int_\varepsilon^{+\infty} \left(\int_\varepsilon^r \frac{r\bar{f}(r)}{q^2\sqrt{r^2-q^2}}\,\mathrm{d}q\right)\mathrm{d}r$$ Deduce that $\forall \varepsilon > 0,\ \int_\varepsilon^{+\infty} \frac{F^{\prime}(q)}{q}\,\mathrm{d}q = -2\varepsilon \int_\varepsilon^{+\infty} \frac{\bar{f}(r)}{r\sqrt{r^2-\varepsilon^2}}\,\mathrm{d}r$.
We consider a function $f$ in $\mathcal{B}_1$ whose partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are in $\mathcal{B}_2$. We set:
$$\forall q \in \mathbb{R}^+,\quad F(q) = \frac{1}{2\pi}\int_0^{2\pi} \hat{f}(q,\theta)\,\mathrm{d}\theta = 2\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r$$
We admit that we can interchange the two integrals and therefore that
$$\forall \varepsilon > 0 \quad \int_\varepsilon^{+\infty} \left(\frac{1}{q^2}\int_q^{+\infty} \frac{r\bar{f}(r)}{\sqrt{r^2-q^2}}\,\mathrm{d}r\right)\mathrm{d}q = \int_\varepsilon^{+\infty} \left(\int_\varepsilon^r \frac{r\bar{f}(r)}{q^2\sqrt{r^2-q^2}}\,\mathrm{d}q\right)\mathrm{d}r$$
Deduce that $\forall \varepsilon > 0,\ \int_\varepsilon^{+\infty} \frac{F^{\prime}(q)}{q}\,\mathrm{d}q = -2\varepsilon \int_\varepsilon^{+\infty} \frac{\bar{f}(r)}{r\sqrt{r^2-\varepsilon^2}}\,\mathrm{d}r$.