Convolution integral problem In the process of solving a certain PDE, I've arrived at a convolution integral:
$$\int_{\mathbb{R}^3} G(x-y) \nabla p(y) dy$$
where $x \in \mathbb{R}^3$, $G(z)=\frac{1}{\| z \|}$ and $p(z) = \frac{z_1}{\| z \|^3}$. Note that $G$ and $p$ both vanish at $\infty$. To avoid computing this gradient, I integrated by parts, moving the $y$ derivative over to $G$. Computing the appropriate derivatives and using the vanishing at $\infty$, I get
$$\int_{\mathbb{R}^3} \frac{x-y}{\| x - y \|^3} \frac{y_1}{\| y \|^3} dy.$$
I'm not sure where to go from here. If it helps any, I already know the final result from another source which is using a different derivation that I don't understand. At the end I should get
$$\frac{2 \pi x}{\| x \|^3} x_1.$$
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\half}{{1 \over 2}}
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 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
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Note that $\ds{\,{\rm p}\pars{y}}$ can be written as
$\ds{\,{\rm p}\pars{y} = -\,\partiald{}{y_{1}}{1 \over \norm{y}}}$

Then,
  \begin{align}&\color{#66f}{\large%
\int_{{\mathbb R}^{3}}\,{\rm G}\pars{x - y}\nabla\,{\rm p}\pars{y}\,\dd y}
=-\int_{{\mathbb R}^{3}}\,{\rm p}\pars{y}\nabla\,{\rm G}\pars{x - y}\,\dd y
\\[5mm]&=\int_{{\mathbb R}^{3}}\,{\rm p}\pars{y}\nabla_{x}
\,{\rm G}\pars{x - y}\,\dd y
=\nabla_{x}\int_{{\mathbb R}^{3}}\,{\rm p}\pars{y}\,{\rm G}\pars{x - y}\,\dd y
\\[5mm]&=-\nabla_{x}\bracks{\int_{{\mathbb R}^{3}}\,{\rm G}\pars{x - y}
\pars{\partiald{}{y_{1}}{1 \over \norm{y}}}\,\dd y}
=\nabla_{x}\bracks{\int_{{\mathbb R}^{3}}{1 \over \norm{y}}\,
\partiald{\,{\rm G}\pars{x - y}}{y_{1}}\,\dd y}
\\[5mm]&=-\nabla_{x}\bracks{\int_{{\mathbb R}^{3}}{1 \over \norm{y}}\,
\partiald{\,{\rm G}\pars{x - y}}{x_{1}}\,\dd y}
\end{align}

$$
\color{#66f}{\large%
\int_{{\mathbb R}^{3}}\,{\rm G}\pars{x - y}\nabla\,{\rm p}\pars{y}\,\dd y}
=-\lim_{d\ \to\ \infty}\nabla_{x}\partiald{}{x_{1}}\bracks{%
\int_{{\mathbb R}^{3} \atop \norm{y}\ <\ d\ >\ \norm{x}}{\,{\rm G}\pars{x - y} \over \norm{y}}\,\dd y}
$$

\begin{align}&\color{#66f}{\large%
\int_{{\mathbb R}^{3}}\,{\rm G}\pars{x - y}\nabla\,{\rm p}\pars{y}\,\dd y}
\\[5mm]&=\!\!-\!\!\lim_{d\ \to\ \infty}\!\!\nabla_{x} \partiald{}{x_{1}}\bracks{%
\int_{0}^{\norm{x}}{1 \over \norm{y}}\,{1 \over
\norm{x}}\,4\pi\norm{y}^{2}\,\dd\norm{y}
+\int_{\norm{x}}^{d}{1 \over \norm{y}}\,
{1 \over \norm{y}}\,4\pi\norm{y}^{2}\,\dd\norm{y}}
\\[5mm]&=-4\pi\lim_{d\ \to\ \infty}\nabla_{x}\partiald{}{x_{1}}\bracks{%
\half\,\norm{x} + d - \norm{x}}
=2\pi\,\nabla_{x}\partiald{\norm{x}}{x_{1}}
=2\pi\,\nabla_{x}{x_{1} \over \norm{x}}
\\[5mm]&=2\pi\pars{%
{\nabla_{x}x_{1} \over \norm{x}} + x_{1}\nabla_{x}{1 \over \norm{x}}}
=2\pi\bracks{{\hat{x}_{1} \over \norm{x}} + x_{1}\pars{-\,{1 \over \norm{x}^{2}}
\,{x \over \norm{x}}}}
=2\pi\,{\norm{x}^{2}\,\hat{x}_{1} - x_{1}x \over \norm{x}^{3}}
\end{align}

$$
\color{#66f}{\large%
\int_{{\mathbb R}^{3}}\,{\rm G}\pars{x - y}\nabla\,{\rm p}\pars{y}\,\dd y}
=\color{#66f}{\large%
2\pi\,{x_{\perp}^{2}\,\hat{x}_{1} - x_{1}x_{\perp} \over \norm{x}^{3}}}\,,\qquad
x_{\perp} \equiv x_{2}\,\hat{x}_{2} + x_{3}\,\hat{x}_{3}
$$
