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I was working on a math assignment and got this problem.

$$\int^\infty_{-\infty}\frac{5x}{1+x^2}dx$$

I know the indefinte integral is $$\frac{-5}{2}\ln|1+x^2|+C$$

Why is it when I looked it up I was told undefined?

Working with the process improper integrals:

$$\lim_{n \to \infty } \frac{-5}{2}\ln|1+n^2|+ \lim_{m \to \infty}\frac{5}{2}\ln|1+m^2|$$

Michael Hardy
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wolfcall
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  • This is an improper integral. What happens when you use the definition for that? – Euler....IS_ALIVE Mar 06 '14 at 23:58
  • I believe you meant to say $5dx$... –  Mar 07 '14 at 00:04
  • Sanath you are right – wolfcall Mar 07 '14 at 00:05
  • The last limit could be simply written as $$\frac{5}{2}\lim_{n \to \infty } \ln\left|\frac{1+n^2}{1+n^2}\right|$$ –  Mar 07 '14 at 00:09
  • Why divide? I thought that needed a subtraction. Also one limit has n the other m – wolfcall Mar 07 '14 at 00:11
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    @SanathDevalapurkar No, the point is that the approaches to $+\infty$ and $-\infty$ are independent, thus the limit doesn't exist, and the improper integral is undefined. If one used symmetric integral limits, one would get the Cauchy principal value, which is $0$. – Daniel Fischer Mar 07 '14 at 00:12
  • @DanielFischer Right. Thanks for the clarification! –  Mar 07 '14 at 00:13
  • @DanielFischer So if I understand you correctly it is undefined because both sides the limit doesn't exist because it continues growing to infinity? – wolfcall Mar 07 '14 at 00:15
  • By approaching $\pm\infty$ in different ways, you can obtain any value (including $\pm\infty$) as a limit, or make it oscillate. Because both ends tend to an infinity, and the signs are different, you get an indeterminate form $(+\infty) + (-\infty)$ as the result. – Daniel Fischer Mar 07 '14 at 00:22

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$$ \int_0^\infty \frac{5x}{1+x^2} \, dx = \infty \text{ and } \int_{-\infty}^0 \frac{5x}{1+x^2}\,dx = -\infty. \tag 1 $$

If you take $$ \lim_{C\to\infty} \int_{-C}^C \frac{5x}{1+x^2}\,dx, \tag 2 $$ you get $0$. But $$ \lim_{C\to\infty} \int_{-C}^{2C} \frac{5x}{1+x^2}\,dx = \frac 5 2 \log_e 4. $$

When the positive and negative parts are both infinite, then "rearranging" the integral can change its value.

This is analogous to the fact that conditionally convergent sums like $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{n}$ (in which the sum of the positive terms and the sum of the negative terms are both infinite) can be made to converge to different things by changing the order of the terms.

The limit in $(2)$ is an example of a "Cauchy principal value" of an integral.

Michael Hardy
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