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I was working on this integral:

$$\int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}$$

Calculations shows that the limits DNE, and therefore the integral diverge. I used Mathematica and found the same result.

But, the integrand is an odd functions, therefore:

$$\forall c \in \Bbb R : \int_{-c}^{+c} \frac{x \, dx}{1+x^2} = 0 $$

So why don't we just say that: $$\int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{c\to\infty} \int_{-c}^{+c} \frac{x \, dx}{1+x^2}=0$$ And the same for any other odd functions?

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    $\begingroup$ what you are looking for is this: en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$
    – ncmathsadist
    Commented Jul 27, 2015 at 21:47
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    $\begingroup$ In a nutshell, because it's equivalent to saying that $-\infty +\infty =0$, which isn't true. $\endgroup$
    – entrelac
    Commented Jul 27, 2015 at 21:55
  • $\begingroup$ in order to show that an improper integral is equal to its cauchy principal value, you should first prove that it's convergent.It's not true in contrary. convergence is the sufficient condition but not the necessary condition $\endgroup$
    – Sepideh Abadpour
    Commented Jul 27, 2015 at 22:06
  • $\begingroup$ My answer below shows that $\displaystyle \vphantom{\frac{\displaystyle \int}{\displaystyle \int}} \lim_{c\to\infty} \int_{-c}^{2c} \dfrac{x\,dx}{1+x^2} = \dfrac 1 2 \log 4$. I don't know why people answer things like this without mentioning things like that. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 27, 2015 at 23:07
  • $\begingroup$ Well, I didn't read before about CPV so I read today and didn't get enough info. about it. What I found is CPV is a method used to evaluate improper integrals when it cannot be evaluated in regular way. So why can't we use CPV in this integral ? $\endgroup$
    – Mohamed Mostafa
    Commented Jul 28, 2015 at 18:09

3 Answers 3

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The definition of the Improper Integral is

$$\begin{align} \int_{-\infty}^{\infty}\frac{x}{1+x^2}dx&\equiv\lim_{L^{-}\to -\infty}\,\,\lim_{L^{+}\to \infty}\int_{L^{-}}^{L^{+}}\frac{x}{1+x^2}dx\\\\ &=\lim_{L^{-}\to -\infty}\,\,\lim_{L^{+}\to \infty} \frac12 \log\left(\frac{(L^{+})^2+1}{(L^{-})^2+1}\right) \end{align}$$

where the integral is defined by taking two separate limits. Inasmuch as this limit does not exist, the integral is undefined.

However, if we interpret the integral as a Cauchy Principal Value, then the upper and lower limits are identical and we have

$$\begin{align} \text{P.V.}\int_{-\infty}^{\infty}\frac{x}{1+x^2}dx&\equiv\lim_{L\to \infty}\int_{-L}^{L}\frac{x}{1+x^2}dx\\\\ &=\lim_{L\to \infty}\frac12 \log\left(\frac{L^2+1}{L^2+1}\right)\\\\ &=0 \end{align}$$

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  • $\begingroup$ I'm not familiar with Cauchy principal value, even I read about it today some articles. So is there One?None? or many solutions for this integral depending on determining the upper and lower limit of integral ? and can you suggest "Simple" and "Intuitive" book or article talking about CPV please? thanks a lot $\endgroup$
    – Mohamed Mostafa
    Commented Jul 28, 2015 at 18:20
  • $\begingroup$ @mohamedmostafa I embedded a link to an article on the CPV. The answer is that the integral is undefined ... Unless one specifies an alternative interpretation of the integral. The CPV interpretation is a common that is used pervasively in applications, including evaluation of integrals via contour integration in the complex plane. $\endgroup$
    – Mark Viola
    Commented Jul 28, 2015 at 18:27
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I would say that, if the integral $$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2} \tag1 $$ does exist, then we have $$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{c\to\infty} \int_{-c}^{+c} \frac{x \, dx}{1+x^2}. \tag2 $$ You have to first prove that the integral in $(1)$ exists to deduce $(2)$.

Think about the following analog situation, you can not assert that $$ (-1)^{\infty}=\lim_{n \to \infty}(-1)^{2n}=1. \tag3 $$ One may recall that

$$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{a \to -\infty}\int_a^c \frac{x \, dx}{1+x^2}+\lim_{b \to +\infty}\int_c^b \frac{x \, dx}{1+x^2},\quad \text{for }\color{red}{\text{any }}c \in \mathbb{R}. $$

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  • $\begingroup$ Your final statement, "One may recall${}\,\ldots$" is incorrect. See my answer. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 27, 2015 at 23:10
  • $\begingroup$ @MichaelHardy I don't see any contradiction between what you wrote and my boxed text above. $\endgroup$
    – Olivier Oloa
    Commented Jul 27, 2015 at 23:16
  • $\begingroup$ The point is that $\lim\limits_{a\to-\infty,\, b\to\infty}$ is not a well defined thing. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 27, 2015 at 23:33
  • $\begingroup$ @MichaelHardy "... is not well defined thing.": so it is difficult to say it "is incorrect" :) Do you prefer $\lim_{a \to -\infty} \lim_{b \to +\infty}$ instead? $\endgroup$
    – Olivier Oloa
    Commented Jul 27, 2015 at 23:38
  • $\begingroup$ If it were $\lim\limits_{a \to -\infty} \lim\limits_{b \to +\infty}$ then it would be $+\infty$ and if it were $\lim\limits_{b \to +\infty} \lim\limits_{a \to -\infty}$ then it would be $-\infty$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 27, 2015 at 23:40
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$$ \int_{-c}^{2c} \frac{x\,dx}{1+x^2} = \frac 1 2 \log\frac{1+4c^2}{1+c^2} \to \frac 1 2 \log 4 \ne 0 \text{ as }c\to\infty. $$

As always with conditionally convergent things, the limit depends on how the bounds approach $\infty$.

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