# Why $\ln (x)-\ln(x)$ as $x$ approaches infinity is not zero?

The questions itself is in "improper integrals" subject.

$$\int \frac{2x}{x^{2}-1}$$

Here is an antiderivative: $$\ln(x^{2}-1)|_{x=-1}^{x=1}$$

If I just place numbers instead of $$x$$ I get $$\ln(0)-\ln(0)$$ which is like writing number-number=zero. The teacher however said that this was a mistake and that it is not based on same principle. why? is it not like substracting a number from itself?

• don't forget $dx$ – J. W. Tanner Jul 19 at 14:01
• Well, what is $ln(0)$? – Sudix Jul 19 at 14:27

It is true that $$\ln(x^2-1)$$ is an antiderivative of $$\dfrac{2x}{x^2-1}$$ on an interval where $$\ln(x^2-1)$$ is defined and differentiable. But unless you want to get into complex values for logarithms of negative numbers, you won't want to use this for $$-1 < x < 1$$. Instead, for $$-1 < x < 1$$ you could take the antiderivative as $$\ln(1-x^2)$$.
Then $$\int_a^b \dfrac{2x}{x^2-1} \; dx = \ln(1-b^2) - \ln(1-a^2)$$ for $$-1 < a < b < 1$$. But you can't use this for $$a=-1$$ and $$b=1$$, because $$\ln(0)$$ is undefined. In fact, you can't take the limit as $$a \to -1$$ and $$b \to 1$$, because $$\ln(z) \to \infty$$ as $$z \to 0+$$. Now you might say, why not take
$$\lim_{b \to 1-}\int_{-b}^b \dfrac{2x}{1-x^2}\; dx = \lim_{b \to 0+} 0 = 0$$ The trouble is that if you do the limit differently, with $$a \ne -b$$, you'll get a different answer. For example, $$\lim_{b \to 1-} \int_{-b^2}^b \dfrac{2x}{1-x^2}\; dx = \lim_{b \to 1-} \ln(1+b^2) = \ln 2$$ So all we can say is $$\int_{-1}^1 \dfrac{2x}{x^2-1}\; dx \ \text{diverges}$$
The map $$x\mapsto\ln(x^2-1)$$ is an antiderivative of $$\dfrac{2x}{x^2-1}$$. But if you replace $$x$$ by to concrete numbers and compute their difference, all you get as number, not an antiderivative of that function. Besides, $$\ln(0)$$ is undefined.
However, it is true that$$\lim_{t\to1^-}\int_{-t}^t\frac{2x}{x^2-1}\,\mathrm dx=0.$$That is so because, for each $$t\in(0,1)$$,$$\int_{-t}^t\frac{2x}{x^2-1}\,\mathrm dx=\ln(t^2-1)-\ln\bigl((-t)^2-1\bigr)=0.$$