How to evaluate limits of the type: $$ \left.[e^{ikx}\Phi(x)]\right|_{x=-\infty}^{x=\infty} $$
Such terms occur while trying to integrate by parts when taking the Fourier Transform of the Poisson Equation. For potentials such as $\Phi(x)=\frac{1}{x}$ this is quite evidently $0$, but I'm not sure how to deal with cases such as the solution to the 2D Poisson Equation: $\Phi(\mathbf{x})=\ln|\mathbf{x}|$. A similar limit occurs from the integral representation of the delta function: $$ \delta(k)=\int_{-\infty}^{\infty}dx e^{ikx} \qquad \implies \qquad ik\delta(k) = \left.[e^{ikx}]\right|_{x=-\infty}^{x=\infty} $$
Let $$f_n(y)=\int_{-n}^n e^{2i\pi yx}dx$$ Then $f_n \to \delta$ in the sense of distributions, which means (by definition) that for all $\phi \in C^\infty_c(\Bbb{R})$ $$\lim_{n\to \infty} \int_{-\infty}^\infty f_n(y) \phi(y)dy = \phi(0)$$ In fact it stays true for any $\phi\in C^1(\Bbb{R})$ which is $L^1$ and whose derivative is $L^1$.
This is how we define and compute the Fourier transforms of things like $1$ and $\log |x|$.