Calculating integrals with asymptotes?

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ One alternative is the Cauchy Principal Value $\ds{\color{#f00}{\mrm{P.V.}}}$: \begin{align} \color{#f00}{\mrm{P.V.}\int_{0}^{2}{\dd x \over \pars{x - 1}^{2}}} & \,\,\,\stackrel{\mbox{def.}}{=}\,\,\, \lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{\dd x \over \pars{x - 1}^{2}} + \int_{1 + \epsilon}^{2}{\dd x \over \pars{x - 1}^{2}}} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\bracks{% {1 \over 1 - \pars{1 -\epsilon}} - {1 \over 1 - 0} + {1 \over 1 -2} - {1 \over 1 - \pars{1 + \epsilon}}} = \lim_{\epsilon \to 0^{+}}\pars{{2 \over \epsilon} - 2} \\[5mm] & = \color{#f00}{\infty} \end{align}

Even the Cauchy Principal Value $\underline{diverges}$.

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John Smith
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Updated on October 02, 2020

Comments

  • John Smith
    John Smith about 3 years

    Find $\displaystyle\int^2_0 \dfrac{1}{(1-x)^2} dx$.

    Is there a way of doing this without considering the asymptote at $x=1$? What if you didn't know at first that there was indeed an asymptote at this point?

    • Winther
      Winther almost 8 years
      If you don't consider the asymptote at $x=1$ you will wrongly conclude that the integral equals $-2$ which is absurd as the integrand is positive. In fact the integral diverges. To properly show this you must split the integral in two $\int_0^1 \frac{1}{(1-x)^2}{\rm d}x + \int_1^2 \frac{1}{(1-x)^2}{\rm d}x$ and analyze both of these integrals. Both must exist if the integral is to exist.
    • okrzysik
      okrzysik almost 8 years
      It doesn't apply in this specific case, but for some divergent integrals you can consider the Cauchy-Principle Value (CPV). The CPV of an integral sort of "dodges" around the fact that the integral is divergent at a given value.