# indefinite integral of 1/x where x is dimensional?

As we know, the integral of $$\frac{1}{x}$$ is $$ln(x)+c$$. Because $$x$$ and $$dx$$ have the same dimension, $$\int\frac{dx}{x}$$ is dimensionless. But my problem is: $$x$$ is dimensional. I've been trained that the natural log of a dimensional quantity is meaningless, and yet here it is.

Furthermore, it seems like this crops up a lot, such as in separable ODE's. E.g. let's say we have $$dx/dt=k x$$, where $$x$$ is dimensional, we obtain $$ln(x)=rt+c$$. Both sides are dimensionless. But then we take the exponential of both sides and get $$x=k e^{rt}$$ -- suddenly both sides have dimensions, and $$k$$ has magically changed from $$e^c$$, which is dimensionless, into dimensional $$k$$, but where did the dimension come from? It seems like every textbook I've read is sloppy about this.

• do you mean the notation is sloppy? The integral is a sum of all the infinitesimal bits of size dx/x, no? – Llouis May 10 at 15:26

$$\ln{(x)} + C = \ln{(Cx)}$$. The dimensions of $$C$$ are those which are needed to make $$Cx$$ dimensionless. :)

• Here we can even talk about the logarithm of a dimension and unit, e.g. $[x] = \log(L),$ $[C] = \log(L^{-1}) = -\log(L).$ – md2perpe May 10 at 16:01

Both members of $$\frac{\dot x}x=r$$ are in $$s^{-1}$$ (speed over distance).

After integration

$$\int\frac{dx}x=\log\left(\frac x{x_0}\right)=r(t-t_0),$$

both members are dimensionless, which allows you to take the exponential:

$$\frac x{x_0}=e^{r(t-t_0)}.$$

Of course,

$$x=x_0e^{r(t-t_0)}$$ are in $$m$$.