Antiderivative of $1/x$, why can $C_1$ and $C_2$ differ?

Question

I haven't taken calculus in a while so please bear with me if this is an elementary question. For the function,

$$ f(x) = 1/x $$

its antiderivative can be written as the piece-wise function,

$$ F(x) = \begin{cases} \ln x + C_1, & \text{if } x > 0 \\ \ln (-x) + C_2, & \text{if } x < 0 \end{cases} $$

Why are the constants defined as $C_1$ and $C_2$ instead of just $C$? That implies to me that they can differ, but I don't understand how they could be unequal. Can someone provide an example of $f(x)$ and $F(x)$ where $C_1 \ne C_2$?

Answer 1

Here is an example:

Let $$F(x) = \begin{cases} \ln x + C_1, & \text{if } x > 0 \\ \ln (-x) + C_2, & \text{if } x < 0 \end{cases}$$

Then derivating, you get $f(x)=\frac{1}{x}$ for $x\neq 0$.

The point of this is that we use the following theorem:

Theorem If $F_1'(x)=F_2'(x)$ on some interval $I$ then, there exists some constant $C$ such that $F_1=F_2+C$.

The theorem is true only over intervals. Since $\frac{1}{x}$ is defined on $(-\infty,0) \cup (0,\infty)$ you have to apply twice the theorem over each of the intervals, getting one (potentially different) constant for each of the two intervals.

Answer 2

It is helpful to think of the space of antiderivatives of a given function as a torsor. See the following: http://math.ucr.edu/home/baez/torsors.html

Answer 3

The function has two separate and independent ''branches'' and the slope $m=F'(x)$ of the two branches does not depends from the added constants but is, in any case $ m=1/x$ . So, yes, the two constants are independent.