# I How do you know when ∞ - ∞ is zero?

1. Oct 14, 2016

### Battlemage!

Suppose I have the following integral:
$$\int_{2}^{∞} \frac{1}{x^2-x}dx$$

Before substituting the boundaries I get
ln|x-1| - ln|x|.

When I put the boundaries in I get

∞ - ∞ - 0 + ln(2) = ln(2). But how do I know those infinities cancel?

Is it because both are infinities of real numbers and therefore are the same size? Is there some other reason?

Thank for any insight.

2. Oct 14, 2016

### olivermsun

You know that $x \ge 2$. So what does that tell you about $\lvert x-2 \rvert$? Can you simplify the difference using this result?

The improper integral (with upper bound $\infty$ means the whole thing is a limit with upper bound $\rightarrow \infty$. So what do you get when you write the whole thing as a limit?

3. Oct 14, 2016

### Staff: Mentor

In general, you don't know. $[\infty - \infty]$ is one of several indeterminate forms, with the brackets there to emphasize that this isn't a number. Other indeterminate forms include $[\frac{\infty}{\infty}], [1^{\infty}]$, and $[\frac 0 0]$. These have to be evaluated on a case-by-case basis.

4. Oct 15, 2016

### andrewkirk

The integral $\int_a^\infty f(x)dx$ doesn't actually mean you evaluate the indefinite integral at infinity. What it strictly means is
$$\lim_{u\to\infty}\int_a^u f(x)dx$$
Similarly the expression
$$\left[\log (x-1)-\log(x)\right]_2^\infty$$
actually means
$$\lim_{u\to\infty}\left[\log (x-1)-\log(x)\right]_2^u$$

Can you work out the following limit?
$$\lim_{u\to\infty}(\log (u-1)-\log(u))$$

5. Oct 15, 2016

### Battlemage!

Not off the top of my head. But I'm thinking if I first combine the logs and then move the log outside the limit I could use L'Hopital's rule.
$$\lim_{u\to\infty}(\log (u-1)-\log(u))$$
$$\lim_{u\to\infty}(\log \frac{u-1}{u})$$
$$\log(\lim_{u\to\infty}(\frac{u-1}{u}))$$
$$\log(\lim_{u\to\infty}(\frac{1}{1}))$$
$$\log(1) = 0.$$

Is that valid?

Anyway I suppose I just took for granted some things about infinity. I have to remember that it is not a number.

6. Oct 15, 2016

### andrewkirk

Yes. Well done.

By the way, that's not l'Hopital's rule you used. L'hopital's rule is about replacing the limit of a ratio by the limit of the ratio of derivatives.

What you used is the theorem that if a function $f$ is continuous then $\lim f(g(x))=f(\lim(g(x))$ provided $\lim(g(x))$ exists, and you applied it with $f=\log$.

7. Oct 15, 2016

### Battlemage!

Hahah well I applied it at this stage:
$$\log(\lim_{u\to\infty}(\frac{u-1}{u}))$$
cause the derivative of u-1 = 1 and the derivative of u = 1, but by the time I got there I did realize there was no point to it. Sadly by then I had already typed it up and didn't feel like deleting.

I was curious, however, if what I did with the limit and the log was valid. I know you can do it for roots, but for logs I was unsure. You have confirmed for me that it is a legitimate move, so I appreciate that. :)