Angular momentum of the electric field of a point-like electric charge and the magnetic field of a monopole I am currently reading "Magnetic Monopoles" of Ya. Shnir. My problem is I can not retrieve a result the author provides in the first chapter of the first part. In this chapter, he studies the non-relativistic scattering of an electric charge on a magnetic one. 
The author writes [p.5, near eq. (1.13)]:

... the appearance of an additional term in the definition of the angular momentum $(1.11)$ originates from a non-trivial field contribution. Indeed, since a static monopole is placed at the origin, its magnetic field is given by $(1.1)$. Then the classical angular momentum of the electric field of a point-like electric charge, whose position is defined by its radius vector $\mathbf{r}$, and the magnetic field of a monopole is a volume integral involving the Poynting vector
\begin{align}
\tilde{\mathbf{L}}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [ \mathbf{E} \times \mathbf{B}\right] d^3r'\tag{L.1}\\& =  - \dfrac{g}{4\pi} \int d^3r' \left( \mathbf{\nabla}'\cdot \mathbf{E}\right) \hat{\bf r}' \tag{L.2}\\ &= -eg\hat{\bf r} \tag{L.3}
\end{align}
where we perform the integration by parts, take into account that the fields
  vanish asymptotically and invoke the Maxwell equation
\begin{equation}\left(\mathbf{\nabla}' . \mathbf{E} \right) = 4 \pi e \delta^{(3)}\left( \mathbf{r} - \mathbf{r}'\right)\end{equation}
  ...

The magnetic field is 
$\mathbf{B} = \dfrac{g}{r^3} \mathbf{r} \tag{1.1}$
The generalised angular momentum is 
$\mathbf{L} = \mathbf{r} \times m\mathbf{v} - eg \hat{\bf r} \tag{1.11}$
The author gives how he got $(L.2)$ from $(L.1)$ but I do not know how to do? Have you any idea?
 A: Answer expected by following author's hints.
\begin{align}
\mathbf{L}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [\mathbf{E} \times \mathbf{B} \right] d^3r'\\
& = \dfrac{1}{4\pi} \int \left [ \left (\mathbf{B.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right) \mathbf{B} \right] d^3r'\\
& = \dfrac{1}{4\pi} \int \left [ \left (\dfrac{g}{r'^3}\mathbf{r'.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right)\dfrac{g}{r'^3} \mathbf{r'} \right] d^3r'\\
& = \dfrac{g}{4\pi} \int \dfrac{1}{r'} \left [\mathbf{E} - \left (\mathbf{E . \hat{r}'} \right) \mathbf{\hat{r}'} \right] d^3r' \\
& = \dfrac{g}{4\pi} \int \left[ \mathbf{E.\nabla'}\right]\mathbf{\hat{r}'} d^3r'.
\end{align}
Or, let $\mathbf{U}$ and $\mathbf{v}$ be arbitrary vectors: 
\begin{equation}
\left [ \mathbf{U.\nabla}\right]\mathbf{v} = \left[\mathbf{U.\nabla}v^i\right]\mathbf{e}_i,
\end{equation}
where $(\mathbf{e}_i)_{1\leq i \leq 3}$ denotes the cartesian basis.
By integrating by parts we have :
\begin{align}
\mathbf{L}_{eg} & = \dfrac{g}{4\pi} \int \left[ \mathbf{E.\nabla'}\right]\mathbf{\hat{r}'} d^3r'\\
& = \dfrac{g}{4\pi} \int \mathbf{E.\nabla'}(\hat{r}'^i) d^3r' \mathbf{e}_i \\
& =  \dfrac{g}{4\pi} \int \left [\mathbf{\nabla' .} \left(\mathbf{E}\hat{r}'^i \right) \mathbf{e}_i - \left (\mathbf{\nabla' . E} \right)\mathbf{\hat{r}}'\right]d^3r'\\
& = \dfrac{g}{4\pi}\ \left[ \oint  \mathbf{\hat{r}'} \left (\mathbf{E.da}\right) - \int \left ( \mathbf{\nabla' . E}\right) \mathbf{\hat{r}'} d^3r' \right].
\end{align}
But the field $\mathbf{E}$ vanishes at infinty so it comes :
\begin{equation}
\mathbf{L}_{eg} = -\dfrac{g}{4\pi} \int \left ( \mathbf{\nabla' . E}\right) \mathbf{\hat{r}'} d^3r'.
\end{equation}
And finally, using the Maxwell equation :$\mathbf{\nabla'.E} = 4\pi e \delta^{(3)}(\mathbf{r} - \mathbf{r'})$, we get the result : 
\begin{equation}
\mathbf{L}_{eg} = -eg\mathbf{\hat{r}}.
\end{equation}
