Fourier transform of the logarithm in three dimensions

Question

I would like to evaluate the (inverse) Fourier transform of $\log(\mathbf q^2+a^2)$ on $\mathbb R^3$, i.e. formally $$ \int \frac{d^3q}{(2\pi)^3}\log(\mathbf q^2+a^2)e^{i\mathbf q\cdot \mathbf x}\,, $$ where $a$ is a positive real constant and $d^3q$ stands for $dq_1 dq_2 dq_3$.

Answer 1

Let us denote $$ \int \frac{d^3q}{(2\pi)^3} \log(\mathbf q^2+a^2) e^{i\mathbf q \cdot \mathbf x}=\mathcal F^{-1}\big[ \log(\mathbf q^2+a^2) \big] \equiv T(\mathbf x)\,. $$ Using the properties of the Fourier transform, $$\begin{aligned} -\mathbf x^2 T(\mathbf x)&=\mathcal F^{-1}\big[\partial_{\mathbf q}^2 \log(\mathbf q^2+a^2) \big] \\ &= 2 \mathcal F^{-1}\big[ (\mathbf q^2+a^2)^{-1}\big]+4a^2\mathcal F^{-1}\big[(\mathbf q^{2}+a^2)^{-2}\big]\\ &=(2-2a \partial_a)\mathcal F^{-1}\big[ (\mathbf q^2+a^2)^{-1}\big]\,. \end{aligned}$$ Now, the inverse Fourier transform of $(\mathbf q^2+a^2)^{-1}$ can be calculated directly: denoting $r\equiv |\mathbf x|$ $$ \int \frac{d^3q}{(2\pi)^3}\frac{e^{i\mathbf q \cdot \mathbf x}}{q^2+a^2}=\frac{1}{2\pi^2r}\int_0^\infty \frac{q\, \sin(qr)}{q^2+a^2} dq =-\frac{1}{4\pi^2 r}\frac{\partial}{\partial r}\int_{-\infty}^{+\infty}\frac{e^{iqr}}{q^2+r^2}dq=\frac{e^{-ar}}{4\pi r}\,, $$ where the last integral can be calculated by standard contour integration methods. Substituting back, we find $$ -r^2 T(r)= \frac{e^{-ar}}{2\pi r}+\frac{a e^{-ar}}{2\pi} $$ whose general solution is $$ T(r)=-\frac{e^{-ar}}{2\pi r^3}-\frac{a e^{-ar}}{2\pi r^2}+C\delta(r^2)\,, $$ where $1/r^3$ and $1/r^2$ are understood as derivatives of the principal value PV$(1/r)$ and $C$ is a constant. However, in $\mathbb R^{3}$, the distribution $\delta(r^2)$ is identically zero: $$ \langle \delta(\mathbf x^2), \varphi(\mathbf x)\rangle = \int d\varphi\, d\theta \,\sin\theta\, r^2 \delta(r^2) \varphi(\mathbf x)=0\,, $$ so the final answer is independent of $C$: $$\boxed{ T(r)=-\frac{e^{-ar}}{2\pi r^3}-\frac{a e^{-ar}}{2\pi r^2} }\,. $$