Integration of Laplacian by parts I'm trying to solve assignment (1.5) in Bellan's "Fundamentals of Plasma Physics" using Fourier transforms, but I'm stuck integrating the Laplacian. Here's the problem:

Equation (1.5) is the following:
$$\nabla^{2} \Phi - \frac{1}{\lambda_D^2} \Phi = \frac{q_T}{\epsilon_0} \delta(\vec r).\tag{1.5}$$
Now if I multiply this with $e^{-i \vec k \cdot \vec r }$ and integrate over $\vec r$ I get simple results for the right-hand side ($-\frac{q_T}{\epsilon_0}$) and the second term on the left-hand side ($\frac{1}{\lambda_D^2} \tilde \Phi (\vec k)$), but can't wrap my head around the first integral. Integrating by parts gives:
$\int \nabla^{2} \Phi e^{-i \vec k \cdot \vec r } d \vec r = \nabla \Phi e^{-i \vec k \cdot \vec r } -  \int \nabla \Phi (-i \vec k) e^{-i \vec k \cdot \vec r } d \vec r $
And using integration by parts on the second term on the right-hand side again gives:
$\int \nabla^{2} \Phi e^{-i \vec k \cdot \vec r } d \vec r = \nabla \Phi e^{-i \vec k \cdot \vec r } - [ \Phi (-i \vec k) e^{-i \vec k \cdot \vec r } - \int \Phi (-i \vec k)^2 e^{-i \vec k \cdot \vec r } d \vec r ] = \nabla \Phi e^{-i \vec k \cdot \vec r } + \Phi i \vec k e^{-i \vec k \cdot \vec r } + k^2 \tilde \Phi (\vec k)$
However, to obtain the correct result (1.43) for the potential in the assignment, I figure that the first two terms should vanish, but I don't know why. I've never integrated the Laplacian, so I could be terribly mistaken in my approach; any help is greatly appreciated!
 A: Integrating by parts once gives:
$\int d^3 r e^{-i\vec k\cdot \vec r}\nabla^2\phi(\vec r) = \int d^3r e^{-i\vec k\cdot \vec r}\vec \nabla\cdot \vec \nabla\phi(\vec r) = \int d^3r \vec \nabla\cdot\left(e^{-i\vec k\cdot \vec r}\vec \nabla \phi\right) +i\vec k\cdot\int d^3re^{-i\vec k\cdot \vec r}\vec \nabla \phi$
Then the first term becomes an integral over the surface (at infinity!):
$$
\int d^3r \vec \nabla\cdot\left(e^{-i\vec k\cdot \vec r}\vec \nabla \phi\right) = \oint \vec dS\cdot \left(e^{-i\vec k\cdot \vec r}\vec \nabla \phi\right)
$$
and if we assume that $\vec \nabla \phi$ goes to zero fast enough at infinity then we can throw this surface term away. (By the way, you can justify this assumption after the fact by Fourier Transforming back to real space to see that $\Phi(\vec r)$ actually does goes to zero at infinity very quickly.)
So, we have (in this case where the field falls off fast enough):
$$
\int d^3 r e^{-i\vec k\cdot \vec r}\nabla^2\phi(\vec r) = +i\vec k\cdot\int d^3re^{-i\vec k\cdot \vec r}\vec \nabla \phi
$$
Integrate by parts one more time (and throw away the surface term again) to find that
$$
\int d^3r e^{-i\vec k\cdot \vec r} \nabla^2\phi(r) = -k^2\phi_k\;.
$$
So, now you know the "rule" $\left(-\nabla^2\right) \to \left(k^2\right)$ when Fourier transforming (for fields that go to zero at infinity faster than $1/r^2$).

In other words, using our rule, we can just read off the answer...
Your Eq (1.5) (which might be missing a minus sign on the RHS?):
$$
\nabla^{2} \Phi - \frac{1}{\lambda_D^2} \Phi = \frac{q_T}{\epsilon_0} \delta(\vec r)
$$
Fourier Transforms to:
$$
-k^2 \Phi_k - \frac{1}{\lambda_D^2} \Phi_k = \frac{q_T}{\epsilon_0}
$$
Or:
$$
\Phi_k = \frac{-q_T}{\epsilon_0\left(k^2 + \lambda_D^{-2}\right)}
$$
