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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?

Michael Hardy
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Mohamed Mostafa
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    what you are looking for is this: https://en.wikipedia.org/wiki/Cauchy_principal_value – ncmathsadist Jul 27 '15 at 21:47
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    In a nutshell, because it's equivalent to saying that $-\infty +\infty =0$, which isn't true. – entrelac Jul 27 '15 at 21:55
  • 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 – Sepideh Abadpour Jul 27 '15 at 22:06
  • 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. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:07
  • 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 ?

     – Mohamed Mostafa Jul 28 '15 at 18:09
  • @MohamedMostafa : One can use CPV here, but it's not the whole story. The integral can take different values if the bounds approach infinity differently. That can happen because $\int_{-\infty}^\infty \left|\frac{x , dx}{1+x^2}\right| = \infty$. If the integral of the absolute value had been finite, then the integral from $-\infty$ to $+\infty$ would be the same finite number regardless of how the bounds approach infinity. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 18:13
  • So the story is this integral can be evaluated as different values depending on how fast its limits approaches infinity ? many possible solutions ? – Mohamed Mostafa Jul 28 '15 at 18:15
  • Correct. See my answer below. – Michael Hardy Jul 28 '15 at 19:09

3 Answers3

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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

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

Mark Viola
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  • 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 – Mohamed Mostafa Jul 28 '15 at 18:20
  • @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. – Mark Viola Jul 28 '15 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

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

Olivier Oloa
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  • Your final statement, "One may recall${},\ldots$" is incorrect. See my answer. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:10
  • @MichaelHardy I don't see any contradiction between what you wrote and my boxed text above. – Olivier Oloa Jul 27 '15 at 23:16
  • The point is that $\lim\limits_{a\to-\infty,, b\to\infty}$ is not a well defined thing. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:33
  • @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? – Olivier Oloa Jul 27 '15 at 23:38
  • 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 $-\infty$. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:40
  • @MichaelHardy Please, how do you define $ \int_{-\infty}^{+\infty} \frac{x , dx}{1+x^2}$? Thanks. – Olivier Oloa Jul 27 '15 at 23:41
  • I don't. But I know how to define the Cauchy principal value and the values of various limits as the bounds approach $\infty$ in various ways. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:43
  • To be clear: $\displaystyle \lim_{a\to,-\infty}\left( \lim_{b\to,+\infty} \int_a^b \frac{x,dx}{1+x^2} \right)$ $\displaystyle = \lim_{a\to,-\infty} \Big( +\infty \Big)$ $=+\infty$. ${}\qquad{}$ – Michael Hardy Jul 27 '15 at 23:45
  • @MichaelHardy I think I know that... But I don't really see the point with what I wrote in my answer. You are telling me that the integral does not exist, which is something I first pointed out... – Olivier Oloa Jul 27 '15 at 23:50
  • You say "One may recall that${},\ldots$" and then say the integral is equal to a certain limit, but in fact it is not. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 00:03
  • @MichaelHardy I'm not sure to understand the point. I have a question, please what does $\int_{-\infty}^{+\infty} \frac{x , dx}{1+x^2} $ means to you? – Olivier Oloa Jul 28 '15 at 00:15
  • It's an indeterminate form. By letting the bounds approach $\pm\infty$ in various different ways, one can make it approach anything in the closed interval $[-\infty,+\infty]$. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 01:10
  • @MichaelHardy So you take two limits (of both bounds) and see if they exist. If you write it down with mathematical symbols, you are not so far from the shortcut $\lim\limits_{a\to-\infty,, b\to\infty}$. But I think a more rigourous way is to consider the form $\lim\limits_{a\to-\infty}\int_a^c \frac{x , dx}{1+x^2}+\lim\limits_{b\to+\infty}\int_c^b \frac{x , dx}{1+x^2}$, for any $c \in \mathbb{R}$. – Olivier Oloa Jul 28 '15 at 01:45
  • $\displaystyle\lim_{a\to-\infty}\int_a^c\frac {x,dx}{1+x^2} \vphantom{\frac{\displaystyle\int}{\displaystyle\int}}$ is equal to $-\infty$ no matter what number $c$ is, and the other limit you give is always $+\infty$. There's nothing more rigorous about what you suggest than about anything else that's been said here. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 18:09
  • So, to understand your answer. What you mean is there is no solution for this integral ( Diverge ) or there are many "possible" solutions depends on assumption of same upper and lower limit (c) or different ?! – Mohamed Mostafa Jul 28 '15 at 18:13
  • @MohamedMostafa Yes, your integral is divergent. This is the usual sense. Now there exist some technics of regularization. The principal Cauchy value is just one of them. Thanks. – Olivier Oloa Jul 28 '15 at 18:46
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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 $\infty$.

Michael Hardy
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