Does $\int_{-\infty}^\infty \frac{1}{x-a}dx$ converge? I don't think this integral converges:
$$\int_{-\infty}^\infty \frac{1}{x-a}dx\qquad x,a\in\mathbb R$$
The integrand is unbounded.
I keep getting undefined natural logarithms and non-existent limits (infinites) whenever I try to work this integral around.
Can you confirm or prove otherwise?
Thank you!
 A: Yes, the integrand is unbounded, but it doesn't follow from that that the integral diverges. In fact, the integral $\int_0^1\frac{\mathrm dx}{\sqrt x}$ converges, although the integrand is unbounded too in this case.
On the other hand, by definition,$$\int_{-\infty}^\infty\frac{\mathrm dx}{x-a}=\int_{-\infty}^a\frac{\mathrm dx}{x-a}+\int_a^\infty\frac{\mathrm dx}{x-a}$$and$$\int_a^\infty\frac{\mathrm dx}{x-a}=\int_a^b\frac{\mathrm dx}{x-a}+\int_b^\infty\frac{\mathrm dx}{x-a}$$for some $b>a$ (the choice of $b$ is irrelevant). But\begin{align}\int_a^b\frac{\mathrm dx}{x-a}&=\lim_{y\to a}\int_y^b\frac{\mathrm dx}{x-a}\\&=\lim_{y\to a}\bigl(\log(b-a)-\log(y-a)\bigr)\\&=\infty\end{align}and\begin{align}\int_b^\infty\frac{\mathrm dx}{x-a}&=\lim_{y\to\infty}\int_b^y\frac{\mathrm dx}{x-a}\\&=\lim_{y\to\infty}\bigl(\log(y-a)-\log(b-a)\bigr)\\&=\infty\end{align}too.
A: It diverges at $x\to+\infty$.  It diverges at $x \to-\infty$ it diverges at $x \to a$.  Summary: it diverges.
Physicists may investigate a "principal value" for this, but in this forum we do not consider that unless the problem specifically mentions it.
