Rewriting an integral This is concerning Poisson's equation with oblique boundary condition (Gilbarg Trudinger p121)
We let $\Gamma(|x-y|)$ denote the fundamental solution to Laplace's equation.  Also, let $x-y^{*} = (x_1-y_1, \cdots ,x_{n-1}-y_{n-1},x_n+y_n)$.  Finally, let  $\zeta = \frac{(x-y^{*})}{|x-y^{*}|}$
I don't understand the computations to get from this 
$$ \Theta = -2b_n\int_0^{\infty}{e^{as}D_n\Gamma(x-y^{*}+\textbf{b}s)ds}$$ where $a\leq 0$
to this
$$ \Theta = -|x-y^{*}|^{2-n}\left( (\frac{2 b_n}{n\omega_n})\int_0^{\infty}e^{a|x-y^{*}|s}\frac{\zeta_n+b_ns}{(1+2({\textbf{$\zeta$}\cdot\textbf{b})s+s^2)^{\frac{n}{2}}}}ds\right)$$
where $\omega_n$ is the volume of the n-ball.  I guess I'm getting stuck in one regard because we only define the fundamental solution for $|x-y|$, and I've never seen something where you are adding a vector inside.  Also, how do we get the additional term in the exponent?  If someone could point me in the right direction, I would appreciate it.  Thanks.
 A: The part about adding a vector inside the argument of $\Gamma$ is not a problem, because $\Gamma$ is a function of a vector, not of a scalar. That is, the definition is
$$
\Gamma(x) = c_n |x|^{2-n},
\qquad x\in\mathbb{R}^{n}\setminus\{0\}
$$
as opposed to 
$$
\Gamma(r) = c_nr^{2-n},
\qquad r>0.
$$
The transformation of the integral is the result of the change of variable where you replace $s$ by $|x-y^*|s$. Let us compute the normal derivative of $\Gamma$ first. We have
$$
\Gamma(x-y^*+bs) = \frac{|x-y^*+bs|^{2-n}}{n(2-n)\omega_n},
$$
and so
$$
D_n\Gamma(x-y^*+bs) 
= \frac{|x-y^*+bs|^{-n}}{2n\omega_n}
\cdot D_n|x-y^*+bs|^2
= \frac{|x-y^*+bs|^{-n}}{2n\omega_n}
\cdot 2(x_n-y^*_n+b_ns).
$$
Now we introduce the new variable $r$ by $s=|x-y^*|r$. Using $r$ and $\zeta$, we can rewrite $D_n\Gamma$ as
$$
D_n\Gamma(x-y^*+bs) 
= \frac{(\zeta_n+b_nr)|x-y^*|}{n\omega_n|\zeta+br|^n|x-y^*|^n}
= \frac{|x-y^*|^{1-n}}{n\omega_n}\cdot\frac{\zeta_n+b_nr}{|\zeta+br|^n}.
$$
We are done once we take into account $ds=|x-y^*|dr$, and
$$
|\zeta+br|^2 = |\zeta|^2 + 2r(\zeta\cdot b) + |b|^2r^2 = 1 + 2r(\zeta\cdot b) + r^2.
$$
