I got stucked in a simple Fourier transform of a vector calculus equation...
Let $\lambda$ and $\mu$ be two scalar fields and $\vec{v}$ be a vector fields. Fourier transform the following equation $$\nabla^2 \vec{v}=\nabla \times (\nabla \lambda \times \nabla \mu)$$
My attempt is:
let $\lambda(x)=\int d\vec{p} \hat{\lambda}_p e^{i\vec{p}\cdot\vec{x}}$ and $\mu(x)=\int d\vec{q} \hat{\mu}_q e^{i\vec{q}\cdot\vec{x}}$
$$\nabla \times \int d\vec{p}\,\, \, (i\vec{p})\hat{\lambda}_p e^{i\vec{p}\cdot\vec{x}} \times \int d\vec{q} \,\,\, (i\vec{q})\hat{\mu}_q e^{i\vec{q}\cdot\vec{x}} $$
$$=\nabla \times \int \int d\vec{p}\,d\vec{q} \,\, (-\vec{p}\times \vec{q}) \hat{\lambda}_p \hat{\mu}_q e^{i(\vec{p}+\vec{q})\cdot\vec{x}}$$
$$-i\int \int d\vec{p}\,d\vec{q} \,\, ((\vec{p}+\vec{q})\times (\vec{p}\times \vec{q})) \hat{\lambda}_p \hat{\mu}_q e^{i(\vec{p}+\vec{q})\cdot\vec{x}} $$
which would vanish since $$(\vec{p}+\vec{q})\times (\vec{p}\times \vec{q})=0$$
Could anyone give a hint on what I did wrong?
Edits: One condition on $\vec{v}$ is that it is divergence-free, i.e. $\nabla\cdot \vec{v}=0$. This might be important but I forgot to state.
The book claims that $$v(\vec{k})=\int \int \,d\vec{k}_1 \,d\vec{k}_2\, \varphi(\vec{k}_1,\vec{k}_2) a^*(k_1)\,a(\vec{k}_2) \, \delta(\vec{k}+\vec{k}_1-\vec{k}_2)$$ where $$\varphi(\vec{k}_1,\vec{k}_2)=\frac{1}{2(2\pi)^{3/2}}[\vec{k}_1+\vec{k}_2-(\vec{k}_1-\vec{k}_2)\frac{k_1^2-k_2^2}{|\vec{k}_1-\vec{k}_2|^2}]$$ and a transition of varialbe is defined as below $$\hat{\mu}(\vec{k})=\frac{a(\vec{k})+a^*(-\vec{k})}{\sqrt{2}},\quad \hat{\lambda}(\vec{k})=\frac{a(\vec{k})-a^*(-\vec{k})}{\sqrt{2}}$$