# (Improper) Integral -2x/(1+x^2) dx from -INF to INF

1. Oct 25, 2013

### s3a

1. The problem statement, all variables and given/known data
Consider the integral

INT -2x/(1+x^2) dx FROM -INF TO INF (The attached TheIntegral.jpg file shows this in a more aesthetically-pleasing manner.)

If the integral is divergent, state that it is so. Otherwise, evaluate the integral.

2. Relevant equations
Integration.
U substitution.

3. The attempt at a solution
I turned the integral the problem gives into:

lim t-> inf INT -2x/(1+x^2) dx FROM -t TO t

Letting u = 1+x^2, du/dx = 2x, dx = du/2x

Lower limit: u = 1 + (-t)^2 = 1 + t^2, Upper limit: u = 1 + (t)^2 = 1 + t^2
lim t-> inf INT -1/u du FROM (1 + t^2) TO (1 + t^2)
lim t-> inf [-ln|u|] FROM (1 + t^2) TO (1 + t^2) = 0

Using Wolfram Alpha, I know the integral is divergent but, when I use Maxima, I get Principal Value: 0 (and, I don't know what means, but it's what I get).

In addition to me agreeing with the value that Maxima gives, I also used another piece of software called Kmplot which is how I got the image I attached in this thread, TheGraph.jpg, which shows a symmetric graph with areas that, to my intuition at least, seem to cancel out.

Could someone please explain to me why this integral is a divergent integral rather than one that converges to a value of 0?

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• ###### TheGraph.jpg
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2. Oct 25, 2013

### szynkasz

Improper integral $\int_{-\infty}^\infty f(x)\,dx$ is converged if both integrals $\int_{-\infty}^af(x)\,dx$ and $\int_a^\infty f(x)\,dx$ are convergent. Cauchy principal value is a limit $\lim\limits_{a\to\infty}\int_{-a}^af(x)\,dx$ and it may exist even if $\int_{-\infty}^\infty f(x)\,dx$ is divergent.

3. Oct 25, 2013

### Staff: Mentor

I posted a reply, but I deleted it and am rethinking what I said...

4. Oct 25, 2013

### s3a

Here is the Wolfram Alpha link.:
http://www.wolframalpha.com/input/?i=integrate+-2x/(1+x^2)+dx+from+-inf+to+inf

Looking at the above link again, it does say that the integral equals to zero below "Cauchy Principal Value:", which is beyond what I'm currently learning, but it also says that the integral does not converge.

According to the answer from the problem I am doing, the integral is in fact diverging (but, I still don't understand why).

5. Oct 25, 2013

### s3a

Szynkasz, that was the information I needed!

Thank you both! :)