# ln(infinity/infinity)

I am taking AP Calculus BC this year, and we are going over improper integrals. I was just doing this integral, and was wondering what exactly the ln(inf/inf) is. Here is my work:

I believe my teacher had said something about how it equals 1 because x-1 and x+1 are of the same degree. Does that make sense? Does L'Hôpital's rule have anything to do with it? I'm pretty sure that's only for limits, but who knows..

• Nitpick: All your integrals are lacking "$dx$". – Hans Lundmark Oct 5 '18 at 6:49
• @HansLundmark ah yes, my bad. If I remember correctly, the 'dx' is the width of each Riemenn area, if you think about it in terms of that. Is that right? – Addison Oct 5 '18 at 13:02
• Yes, that's right. And that's why the integrals look a bit funny without it. – Hans Lundmark Oct 5 '18 at 13:13

The antiderivative is right. But you can’t “plug in $$\infty$$”: you need to compute a limit: $$\lim_{x\to\infty}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|= \lim_{x\to\infty}\frac{1}{2}\ln\left|\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\right|= \frac{1}{2}\ln 1=0$$ Similarly for the lower bound: $$\lim_{x\to1^+}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|=-\infty$$ The integral is not convergent.

• I see, so infinity cannot be treated as a number. It must be treated as a limit, correct? – Addison Oct 4 '18 at 23:54
• @Addison: correct – Clayton Oct 5 '18 at 0:05