Multidimensional Inverse Fourier transform of a function

Question

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.

Answer 1

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 $$ \mathbf{k}=(k\sin\phi\cos\theta,k\sin\phi\sin\theta,k\cos\phi),\quad \mathrm{d}_3k=k^2\sin\phi\,\mathrm{d}\theta\,\mathrm{d}\phi\,\mathrm{d}k. $$

Answer 2

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: $$F(\mathbf x)=\int\frac{e^{i\,\mathbf k\cdot\mathbf x}}{A\,k^2+s^{2\nu}}\,\mathrm{d}_3k=\int \frac{e^{ikr\cos\theta}}{A\,k^2+s^{2\nu}}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\, k^2\mathrm{d}k$$ $$=\left(\int_0^{2\pi}\mathrm{d}\phi\right)\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\int_0^{\pi} e^{ikr\cos\theta}\sin\theta\,\mathrm{d}\theta\right)\,\right]\, k^2\mathrm{d}k$$ $$=2\pi\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\int_{-1}^{1} e^{ikrt}\,\mathrm{d}t\right)\,\right]\, k^2\mathrm{d}k$$ $$=2\pi\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(\frac{2\sin(kr)}{kr}\right)\,\right]\, k^2\mathrm{d}k$$ $$=\frac{2\pi}{ir}\int_0^\infty\left[ \frac{1}{A\,k^2+s^{2\nu}}\left(e^{ikr}-e^{-ikr}\right)\,\right]\, k\mathrm{d}k$$ $$=\frac{2\pi}{ir}\int_{-\infty}^\infty\frac{e^{ikr}}{A\,k^2+s^{2\nu}}\, k\mathrm{d}k$$ $$=\frac{2\pi}{ir}\frac{\partial}{i\partial r}\int_{-\infty}^\infty \frac{e^{ikr}}{A\,k^2+s^{2\nu}}\, \mathrm{d}k$$