Blackholeinside said:
Since it is infinite it will be infinite between any point outside the black hole and a radial point on the event horizon.
Please show your work. Or no, I have a better idea; I'll show you the correct calculation.
As Nugatory says, Schwarzschild exterior coordinates don't actually cover the horizon, so the best way to really make this calculation easy is to use other coordinates that do, such as Painleve coordinates. But I'll use a different approach, which is to compute the radial distance between two points, one of which is very close to the horizon, in Schwarzschild coordinates, and then take the limit as the radial coordinate of the lower point goes to the radial coordinate of the horizon.
Since the only coordinate that will vary is ##r## (a purely radial line in a surface of constant coordinate time), the line element is
$$
ds^2 = \frac{1}{1 - 2M / r} dr^2
$$
So the radial distance is the integral of ##ds = \sqrt{ds^2}##, or
$$
s = \int ds = \int_{\rho}^{R} \sqrt{\frac{r}{r - 2M}} dr
$$
where ##r = \rho## is the inner point (the one whose radial coordinate we'll let go to ##2M##) and ##r = R## is the outer point, the one we'll hold constant. Looking up the antiderivative in a table of integrals (for example, see
here), we obtain
$$
s = \left[ \sqrt{r \left( r - 2M \right)} + 2M \ln | \sqrt{r} + \sqrt{r - 2M} | \right]_{\rho}^{R}
$$
This is easily evaluated (I have rearranged the log terms to make things look a bit neater, and removed the absolute value signs since everything inside them is nonnegative):
$$
s = \sqrt{R \left( R - 2M \right)} - \sqrt{\rho \left( \rho - 2M \right)} + 2M \ln \frac{\sqrt{R} + \sqrt{R - 2M}}{\sqrt{\rho} + \sqrt{\rho - 2M}}
$$
We can now take the limit as ##\rho \rightarrow 2M##, and we get an obviously finite answer:
$$
s = \sqrt{R \left( R - 2M \right)} + 2M \ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)
$$
Note, also, that as ##R \rightarrow 2M##, ##s## goes to zero, as it should.