Green's Function of Bi-Laplacian in 3D So I am trying to show that for the equation $\nabla^4G(\boldsymbol{x,x_0})=\delta(\boldsymbol{x,x_0})$ in 3D that $G(\boldsymbol{x,x_0})=-\frac{|x-x_0|}{8\pi}$. I Fourier Transformed the equation and got to the line: $$G^*(\boldsymbol{k})=\int_{V}\nabla^4G(\boldsymbol{x-x_0})e^{-2\pi i\boldsymbol{k\cdot x}}d^3\boldsymbol{x}=\bigg(\dfrac{1}{2\pi|\boldsymbol{k}|}\bigg)^4$$
But then when I try to invert by converting into spherical coordiantes, I end up getting a non-converging integral $$G(\boldsymbol{x,x_0})=-\dfrac{1}{8\pi^4} \int_0^\infty 
\dfrac{\sin(2\pi k \boldsymbol{r})}{k^3 |\boldsymbol{r}|}\,dk\ $$ where $\boldsymbol{r}=\boldsymbol{x-x_0}$.
Any advice for how to proceed from here or what error I may have made getting to this line?
 A: We can, indeed, differentiate distributions.
Let $u\in \mathcal{D}(\mathbb{R}^3),$ and call $v=|x|^\alpha$ ($\alpha$ to-be-determined organically; it will turn out to be $1$) and write $\Omega_\epsilon=\mathbb{R}^3\setminus B_\epsilon.$ Then, 
\begin{align*}
\langle \Delta^2 |x|^\alpha,u\rangle=\langle |x|^\alpha,\Delta^2 u\rangle=\lim_{\epsilon\rightarrow 0} \int\limits_{\Omega_\epsilon} \Delta^2 u|x|^\alpha\ dx=\lim_{\epsilon\rightarrow 0} \int\limits_{\Omega_\epsilon} \left(\Delta^2 u|x|^\alpha -u\Delta^2 |x|^\alpha\right)\ dx,
\end{align*} choosing $\alpha$ so that $\Delta^2|x|^\alpha=0$ for all $x\neq 0.$ One can use Green's theorem to show that $$\int\limits_{\Omega_\epsilon} v\Delta^2 u\, dx-\int\limits_{\Omega_\epsilon} u\Delta^2 v\, dx=\int\limits_{\partial\Omega_\epsilon} \left(v(\partial_\nu \Delta u)-(\partial_\nu v)\Delta u+(\partial_\nu u)\Delta v-u(\partial_\nu\Delta v)\right)\, ds,$$ where $\nu$ is the unit normal on $\partial\Omega_\epsilon=\partial B_\epsilon.$ Applying this to our above quantity with $u$ and $v$ as labeled at the beginning and changing the Laplacian to spherical coordinates, we get that 
$$\langle \Delta^2 |x|^\alpha,u\rangle=\lim_{\epsilon\rightarrow 0}\int\limits_{\partial B_\epsilon} \left(\epsilon^\alpha (\partial_\nu\Delta u)-\alpha\epsilon^{\alpha-1}\Delta u+(\partial_\nu u)(\alpha(\alpha+1)\epsilon^{\alpha-2})-u(\alpha(\alpha+1)(\alpha-2)\epsilon^{\alpha-3})\right)\, ds.$$ We need that this limits to $u(0)$ (times a constant). Note that the area of the region of integration is $\epsilon^{2}\text{Area}(S^2).$ So, we will need that $\alpha=1,$ as it will satisfy all of the given criteria that we've listed. Hence, only the last term survives in the limit, and it tends to $-8\pi u(0)=-8\pi\langle \delta ,u\rangle$. That is,  $$\Delta^2\left(-\frac{|x|}{8\pi}\right)=\delta,$$ as desired.
The same procedure gives the same analogous result when considering the Laplacian. You need to be careful when differentiating; you will not just get $0$. 
