Any meaning for an incorrectly calculated improper integral? In my introductory calculus class we are looking at improper integrals. Specifically, ones that involve infinite discontinuities in the interval of integration. For example, the integral $$\int_{0}^{3} \frac{dx}{x-1} $$ is clearly improper since the numerator of $\frac{1}{x-1}$ is zero when $x=1$ and it is easy to see that it diverges to $-\infty$. However, I am interested if there is any meaning to the answer we obtain if we evaluate this integral incorrectly. By "incorrectly" I mean in this manner:
$$\int_{0}^{3} \frac{dx}{x-1} = \ln|x-1| \Big|_0^3 = \ln2-\ln1 = \ln2$$
Is there any meaning to this at all? It seems (at least to me) we cant give this any physical meaning but is there a more abstract way we can assign meaning to it or is this equivalent to asking if $2+2=3$ has any meaning.
 A: We can assign a meaning to the improper integral in terms of its Cauchy Principal Value.  Then we write
$$\begin{align}
\text{PV}\left(\int_0^3\frac{1}{x-1}\,dx\right)&=\lim_{\epsilon \to 0^+}\left(\int_0^{1-\epsilon}\frac{1}{x-1}\,dx+\int_{1+\epsilon}^3\frac{1}{x-1}\,dx\right)\\\\
&=\lim_{\epsilon \to 0^+}\left(\int_{-1}^{-\epsilon}\frac{1}{x}\,dx+\int_{\epsilon}^2\frac{1}{x}\,dx\right)\\\\
&=\lim_{\epsilon\to 0^+}(\log(\epsilon)+\log(2)-\log(\epsilon))\\\\
&=\log(2)
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{P.V.}\int_{0}^{3}{\dd x \over x - 1} & =
\mrm{P.V.}\int_{-1}^{2}{\dd x \over x} =
\underbrace{\mrm{P.V.}\int_{-1}^{1}{\dd x \over x}}_{\ds{=\ 0}}\ +\
\underbrace{\int_{1}^{2}{\dd x \over x}}_{\ds{=\ \ln\pars{2}}} =
\bbx{\ds{\ln\pars{2}}}
\end{align}
