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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:

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

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This is an improper integral. What happens when you use the definition for that? –  Euler....IS_ALIVE Mar 6 at 23:58
    
I believe you meant to say $5dx$... –  SDevalapurkar Mar 7 at 0:04
    
Sanath you are right –  wolfcall Mar 7 at 0:05
    
The last limit could be simply written as $$\frac{5}{2}\lim_{n \to \infty } \ln\left|\frac{1+n^2}{1+n^2}\right|$$ –  SDevalapurkar Mar 7 at 0:09
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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 7 at 0:12

1 Answer 1

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

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.

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