Integration by Parts for a Line Integral of Fundamental Singularities of Fluid Flow I am reading the following article on a fluid dynamics problem which can be found here https://arxiv.org/pdf/cond-mat/0506212.pdf.  In Appendix C, the author takes the following equation
$(1 - 2 \lambda) \frac{\partial}{\partial z} \textbf{w} (\textbf{r}, \lambda) = 2h^2 \textbf{G}^D_z - 2h \textbf{G}^{SD}_{z,z} - 2 \lambda \textbf{w}_b (\text{r}, \lambda) $
where $\textbf{G}^D_z $ and $\textbf{G}^{SD}_{z,z}$ are derivatives of functions of $\textbf{r}$ defined in the text (singularities of Stokes flow), and uses integration by parts plus the fact that
$\Bigg( 1 - 2 \lambda \frac{\partial}{\partial z}\Bigg) \hat{\textbf{w}}_b (\textbf{r},\lambda) = \hat{\textbf{w}}_b(\textbf{r},0) $
and 
$\textbf{w}_b (\textbf{r}, 0) = -2h \textbf{G}^D_z(\textbf{r})$
to arrive at the integral
$\textbf{w} (\textbf{r}, \lambda) = \frac{h}{\lambda} \int^{\infty}_0 e^{-s/2 \lambda} ((h +s)\textbf{G}^D_z  - \textbf{G}^{SD}_{z,z}) ( \textbf{r} + (h+s) \textbf{e}_z) ds.$
I am unclear where the exponential comes from in this integral.  Also how is the equation for the Fourier transformed quantity $\hat{\textbf{w}}_b(\textbf{r},\lambda) $ being used?  Are we assuming the same relation holds if we invert the quantities back to real space?
 A: I figured out bits of it so this is a partial answer. The exponential term is linked to the inverse of the differential operator $\left(1-2\lambda \dfrac{\partial}{\partial z}\right)$. Let me call the right-hand-side of the PDE $F(\mathbf{r},\lambda)$ and . We are solving 
\begin{align*}
\left(1 - 2\lambda\dfrac{\partial}{\partial z}\right)\mathbf{w} & = F(\mathbf{r},\lambda) \\
\dfrac{\partial\mathbf{w}}{\partial z} - \frac{\mathbf{w}}{2\lambda} & = -\frac{F(\mathbf{r},\lambda)}{2\lambda} \\
\mathbf{w}(\mathbf{r},\lambda) & = C(x,y)e^{z/2\lambda} + \int_z^\infty e^{(z-t)/2\lambda}\frac{F(r|_{z=t},\lambda)}{2\lambda}\, dt
\end{align*} 
where the last step follows from solving the first order PDE in $\mathbf{w}$ using integrating factor. I don't know what the range of $z$ is but they have $\infty$ on their expression and I simply follow them by integrating from $z$ to $\infty$. Now, $C(x,y)\equiv 0$ in order for $\mathbf{w}$ to be bounded as $z\longrightarrow\infty$. Substituting $F$, we get 
\begin{align*}
\mathbf{w}(\mathbf{r},\lambda) & = \int_z^\infty e^{(z-t)/2\lambda}\frac{F(r|_{z=t},\lambda)}{2\lambda}\, dt \\
& = \frac{1}{2\lambda} \int_z^\infty e^{(z-t)/2\lambda}\Big[(2h^2\mathbf{G}_z^D - 2h\mathbf{G}_{z,z}^{SD})(\mathbf{r}|_{z=t}) - 2\lambda\mathbf{w}_b(\mathbf{r}|_{z=t},\lambda)\Big]\, dt \\
& = \frac{h}{\lambda} \int_z^\infty e^{(z-t)/2\lambda}\Big[(h\mathbf{G}_z^D - \mathbf{G}_{z,z}^{SD})(\mathbf{r}|_{z=t}) - \frac{\lambda}{h}\mathbf{w}_b(\mathbf{r}|_{z=t},\lambda)\Big]\, dt. 
\end{align*}
To make the lower limit of the integral to be 0, simply make the change of variable $-s = z - t$. We then get 
\begin{align*}
& \frac{h}{\lambda} \int_z^\infty e^{(z-t)/2\lambda}\Big[(h\mathbf{G}_z^D - \mathbf{G}_{z,z}^{SD})(\mathbf{r}|_{z=t}) - \frac{\lambda}{h}\mathbf{w}_b(\mathbf{r}|_{z=t},\lambda)\Big]\, dt \\
& = \frac{h}{\lambda} \int_0^\infty e^{-s/2\lambda}\Big[(h\mathbf{G}_z^D - \mathbf{G}_{z,z}^{SD})(\mathbf{r}|_{z=z+s}) - \frac{\lambda}{h}\mathbf{w}_b(\mathbf{r}|_{z=t},\lambda)\Big]\, ds.
\end{align*}
Now, I thought $\mathbf{r} = (x,y,z)$ which means $\mathbf{r}|_{z = z+s} = \mathbf{r}+s\mathbf{e_z}$ which is not what they have. But if I take $\mathbf{r} = (x,y,z-h)$ (from page 4, the line just before equation 9), then 
$$ \mathbf{r} + (h+s)\mathbf{e_z} = (x,y,z+s) = (x,y,z)\Big|_{z=z+s}. $$
So I suspect that $\mathbf{r}$ varies from line to line but maybe you can figure out why. 
We now deal with the term $\mathbf{w}_b(\mathbf{r},0)$. Solve the first-order PDE for $\hat{\mathbf{w}}_b$ analogously to how we solve for $\mathbf{w}$:
\begin{align*}
\hat{\mathbf{w}}_b(\mathbf{r},\lambda) & = \frac{1}{2\lambda}\int_z^\infty \hat{\mathbf{w}}_b(\mathbf{r}|_{z=t},0)\, dt
\end{align*}
Then invert the Fourier transform: 
$$ \mathbf{w}_b(\mathbf{r},\lambda) = \mathcal{F}^{-1}\left(\frac{1}{2\lambda}\int_z^\infty \hat{\mathbf{w}}_b(\mathbf{r}|_{z=t},0)\, dt\right) $$
This is triple integral in $k_1, k_2, t$ and we are not integrating over a rectangle, it seems natural to change the order of integration so that $\mathcal{F}^{-1}$ is composed with $\hat{\mathbf{w}}_b(\mathbf{r},0)$ and you get just $\mathbf{w}_b(\mathbf{r},0)$; this is where you used the identity 
$$ \mathbf{w}_b(\mathbf{r}, 0) = -2h\mathbf{G}_z^D(\mathbf{r}). $$ 
Finally, this leads to integrating $\mathbf{w}_b(\mathbf{r},0)$ (Which is an integral) in the expression of $\mathbf{w}(\mathbf{r},\lambda)$ and it is clear that you would want to integrate by parts now, but I get 0 for some reason when I do this so I must be doing something wrong........ 
