I How do you know when ∞ - ∞ is zero?

[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.

 

[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?

 

[Mark44]

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?

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.

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

Thank for any insight.

 

[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?
$$\lim_{u\to\infty}(\log (u-1)-\log(u))$$

 

[Battlemage!]

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?
$$\lim_{u\to\infty}(\log (u-1)-\log(u))$$

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.

 

[andrewkirk]

Is that valid?

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 $f=\log$.

 

[Battlemage!]

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 $f=\log$.

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. :)