Multidimensional Inverse Fourier transform of a function I am studying a paper about transport processes. 
I'm having a difficulty understanding one derivation of an inverse  Fourier transform of a function.
The Fourier transform is defined as
$$
 f(k)= \int_{R^3} e^{- i \ k  \cdot x} f(x) dx_3 \quad 
 \div \quad f(x)= \frac{1}{(2 \pi)^3 } \int_{R^3} e^{  i \ k  \cdot x} f(k) dk_3 \quad 
$$
I am reproducing here the derivation in the paper:

I don't understand the transition from row 1 to row 2.
 A: In spherical coordinates 
$$\mathbf{k}=(k\cos\phi\sin\theta,k\sin\phi\sin\theta,k\cos\theta),\quad \mathrm{d}_3k=\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\,k^2\mathrm{d}k.$$
The integration for $\phi,\theta$ can be carried out explicitly. Here are the details:
$$F(\mathbf x)=\int\frac{e^{i\,\mathbf k\cdot\mathbf x}}{A\,k^2+s^{2\nu}}\,\mathrm{d}_3k=\int \frac{e^{ikr\cos\theta}}{A\,k^2+s^{2\nu}}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\, k^2\mathrm{d}k$$
$$=\left(\int_0^{2\pi}\mathrm{d}\phi\right)\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\int_0^{\pi} e^{ikr\cos\theta}\sin\theta\,\mathrm{d}\theta\right)\,\right]\, k^2\mathrm{d}k$$
$$=2\pi\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\int_{-1}^{1} e^{ikrt}\,\mathrm{d}t\right)\,\right]\, k^2\mathrm{d}k$$
$$=2\pi\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\frac{2\sin(kr)}{kr}\right)\,\right]\, k^2\mathrm{d}k$$
$$=\frac{2\pi}{ir}\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(e^{ikr}-e^{-ikr}\right)\,\right]\, k\mathrm{d}k$$
$$=\frac{2\pi}{ir}\int_{-\infty}^\infty\frac{e^{ikr}}{A\,k^2+s^{2\nu}}\, k\mathrm{d}k$$
$$=\frac{2\pi}{ir}\frac{\partial}{i\partial r}\int_{-\infty}^\infty  \frac{e^{ikr}}{A\,k^2+s^{2\nu}}\, \mathrm{d}k$$
A: It is a change to spherical coordinates. The function
$$
f(\mathbf{k})=\frac{1}{A\,k^2+s^{2\nu}}
$$
is radial. Therefore, its inverse Fourier transform is also radial. This means that
$$
\widetilde{u^{(3)}}(\mathbf x)=\widetilde{u^{(3)}}(0,0,r),
$$
where $r=(x_1^2+x_2^2+x_3^2)^{1/2}$. Then
$$
\int e^{i\,\mathbf k\cdot\mathbf x}\,f(\mathbf k)\,\mathrm{d}_3k=\int e^{ik_3r}\,f(\mathbf k)\,\mathrm{d}_3k.
$$
No change to spherical coordinates
$$
\mathbf{k}=(k\sin\phi\cos\theta,k\sin\phi\sin\theta,k\cos\phi),\quad \mathrm{d}_3k=k^2\sin\phi\,\mathrm{d}\theta\,\mathrm{d}\phi\,\mathrm{d}k.
$$
