Question

Definite intergral problem with infinities as boundaries.

Answer 1

I'll type it out here.

Answer 2

\[\int\limits_{-\infty}^{\infty}\frac{ x }{ x^{2}+1}\]

Answer 3

Break it up first (excuse the notation):
\[\int_{-\infty}^0+\int_0^\infty\]
Then write it as a limit:
\[\lim_{a\to-\infty}\int_a^0+\lim_{b\to\infty}\int_0^b\]
A substitution works, let \(u=x^2+1\).

Answer 4

I tried that and just got infinity for both parts. Does that just mean the intergral diverges?

Answer 5

\[\begin{align*}\lim_{a\to-\infty}\int_a^0\frac{x}{x^2+1}~dx&=\lim_{a\to-\infty}\int_{a^2+1}^1\frac{du}{u}\\\\\\
&=\lim_{a\to-\infty}\left[~\ln\left|u\right|~\right]_{a^2+1}^1\\\\\\
&=\lim_{a\to-\infty}\ln\left|1\right|-\ln\left|a^2+1\right|\\\\
&=0-\lim_{a\to-\infty}\ln(a^2+1)\\\\
&=-\lim_{a\to-\infty}\ln(a^2+1)
\end{align*}\]
You're right that this limit is infinite, but let's look at the other integral first:
\[\begin{align*}\lim_{b\to\infty}\int_0^b\frac{x}{x^2+1}~dx&=\lim_{b\to\infty}\int_1^{b^2+1}\frac{du}{u}\\\\\\
&=\lim_{b\to\infty}\left[~\ln\left|u\right|~\right]_1^{b^2+1}\\\\\\
&=\lim_{b\to\infty}\ln(b^2+1)-0\\
&=\lim_{b\to\infty}\ln(b^2+1)
\end{align*}\]
And again, this limit is infinite. Which means
\[\int_{-\infty}^\infty \frac{x}{x^2+1}~dx=\infty-\infty\]
However, this is an indeterminate form; \(\infty-\infty\) is not necessarily 0.

Answer 6

And just because both are infinite does not mean \(\infty -\infty\) is not finite.

Answer 7

Wow, thank you! But when I looked at the graph of this functions I noticed it was symmetrical about the y axis but the left half is under the x axis and the right side is above it. So, shouldn't the limit be infinity plus infinity?

Answer 8

The area to the left of the y-axis would be negative because it's under the x-axis. It happens to be infinite, so you still have \(\color{red}\infty\color{blue}{-\infty}\), where the red is the positive area (to the right) and the blue is the negative (to the left).

Answer 9

Oh, ok. I see. So...What would I do from there? Can anything be done of does the limit not exist?

Answer 10

This looks like a problem involving L'Hopital's rule. Just wondering how one would set it up, since we have two different limiting variables, \(a\) and \(b\)...

Answer 11

Well, according to WolframAlpha, the integral diverges, so your intuition that it diverges is correct; if one integral diverges, that appears to be enough for the initial integral to diverge.

Answer 12

Ah, alright! Well thank you, your incredibly helpful.

Answer 13

You're*