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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.

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We can assign a meaning to the improper integral in terms of its Cauchy Principal Value. Then we write

$$\begin{align} \text{PV}\left(\int_0^3\frac{1}{x-1}\,dx\right)&=\lim_{\epsilon \to 0^+}\left(\int_0^{1-\epsilon}\frac{1}{x-1}\,dx+\int_{1+\epsilon}^3\frac{1}{x-1}\,dx\right)\\\\ &=\lim_{\epsilon \to 0^+}\left(\int_{-1}^{-\epsilon}\frac{1}{x}\,dx+\int_{\epsilon}^2\frac{1}{x}\,dx\right)\\\\ &=\lim_{\epsilon\to 0^+}(\log(\epsilon)+\log(2)-\log(\epsilon))\\\\ &=\log(2) \end{align}$$

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  • $\begingroup$ Like OP I'm also interested in this topic. However, I can only find a definition of the Cauchy Principal Value rather than meaning/interpretation/use. Could you shed some light on the purpose of this value? $\endgroup$ – Hugh Mar 7 '17 at 23:56
  • $\begingroup$ @ZaidAlyafeai Thank you; much appreciative! And yes, contour integration sometimes involves a principal value integration. $\endgroup$ – Mark Viola Mar 7 '17 at 23:58
  • $\begingroup$ Pretty strange that was mentioned in an introductory calculus course. $\endgroup$ – Zaid Alyafeai Mar 7 '17 at 23:59
  • $\begingroup$ @ZaidAlyafeai The Principal Value was not mentioned in my course but I was inspired to ask this question since my textbook gave this as an example of what not to do when integrating improper integerals. $\endgroup$ – 1730 Mar 8 '17 at 0:29
  • $\begingroup$ @1830, aha, makes sense now! $\endgroup$ – Zaid Alyafeai Mar 8 '17 at 0:34
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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