Proper time of flight Schwarzschild metric being finite With studying Schwarzschild metric geodesics one can easily come up with the following differential equation
\begin{equation}
\dfrac{dr}{d\tau} = - \sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)}
\end{equation}
which relates the radial coordinate and the proper time outside the event horizon $r_H=2GM$ (I'm using, of course, $c=1$).
Imagine that an observer is falling from the initial position $r_0=2GM+h$ (with $h>0$ in order to be outside the event horizon) all the way down to the event
horizon $r_H=2GM$. The proper time of flight is, thanks to the differential equation given above, by
\begin{equation}
\Delta \tau= \int_{\tau_0}^{\tau_H}d\tau = \int_{r_0=2GM+h}^{r_H=2GM} -\;\dfrac{dr}{\sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)}} 
\end{equation}
However, the coordinate time of flight is given by the relation $dt=\frac{C}{(1-2GM/r)}d\tau$ (where $C>0$, see for example http://gfm.cii.fc.ul.pt/events/lecture_series/general_relativity/gfm-general_relativity-lecture4.pdf), resulting
\begin{equation}
\Delta t= \int_{t_0}^{t_H}dt = \int_{r_0=2GM+h}^{r_H=2GM} -\; \frac{C}{1-\dfrac{2GM}{r}}\;\dfrac{dr}{\sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)}} 
\end{equation}
From this (see for example the reference given above) it is said that ''
It is easy to see, by direct evaluation of the integrals, that the proper time of flight is finite, while
the coordinate time of flight is infinite, the orbit observer will never see the infalling observer reach the event horizon, except asymptotical''.
I've tried to calculate the first integral using Mathematica and the result I get is absurd, giving a complex proper time. Here is the code I used:
Integrate[1/Sqrt[C^2 - (1 - 2*G*M/r)], {r, 2*G*M, 2*G*M + h}]  
And the output is:
ConditionalExpression[(G M (2 Sqrt[C^2 - C^4] - I Log[2] - I Log[((-I + 2 I C^2 + 2 Sqrt[C^2 - C^4]) G M)/Sqrt[1 - C^2]]))/(1 - C^2)^(3/2) + (-Sqrt[1 - C^2] (h + 2 G M) Sqrt[C^2 - h/(h + 2 G M)] + I G M Log[2] + I G M Log[(-I h + I C^2 h - I G M + 2 I C^2 G M + Sqrt[1 - C^2] h Sqrt[C^2 - h/(h + 2 G M)] + 2 Sqrt[1 - C^2] G M Sqrt[C^2 - h/(h + 2 G M)])/Sqrt[1 - C^2]])/(1 - C^2)^(3/2), ((G M)/h != 0 && Re[(GM)/h] >= 0) ||Re[(G M)/h] < -(1/2) || (G M)/h \[NotElement] Reals]
What I am missing up here? Thanks
 A: I also do not quite know how this is "easy to see" from that.
However, you can solve the integral for proper time by first noting that for $C^2=\text{const.}$ we have (with $r_s = 2GM$)
$$C^2 = 1 - \frac{r_s}{r_0} $$
which follows from plugging in $(4.2)$ into $(4.1)$ and considering an infalling particle which is at rest at $r = r_0$, meaning $\dot{r}\vert_{r=r_0} = 0$.
This you can use to write your first equation as
$$\dot{r} = -r_s^{1/2} \left(\frac{r_0 - r}{r_0 r}\right)^{1/2}$$
From this, the proper time integral will turn out to be
$$\Delta \tau = -r_s^{-1/2} \int_{r_0}^{r_s} \left(\frac{r r_0}{r_0 - r}\right)^{1/2}\text{d}r $$
which you can solve using Mathematica or also by introducing the parametrization
$$r(\eta) = \frac{r_0}{2} (1+\cos\eta)$$
with $\eta \in [0,\pi]$.
A: 
the result I get is absurd, giving a complex proper time.

Your result has I in it, which I assume is Mathematica's notation for $i=\sqrt{-1}$, but that doesn't mean the result is complex. It's common with certain techniques of integration to get results that look complex but actually turn out to be real when you evaluate them.
You're making your results unnecessarily complicated by taking $G$ and $M$ as parameters. When doing this sort of integral, the thing to do is to convert as many variables as possible to unitless quantities, which generally has the effect of eliminating unnecessary constants like these. Here you want to change to the variable $x=r/2GM$. Don't just feed it into the Computer Algebra System (CAS) without doing this prep. CAS's are stupid and produce complicated output, so give it a fighting chance by doing the same setup you would normally do if working without a CAS.
I use the open-source CAS maxima, rather than mathematica. It has no problem producing a result with no explicit $i$'s in it.
$ maxima -q --batch-string="assume(h>0 and (c^2-1)<0 and c>0); integrate(-(c^2-(1-1/x))^(-1/2),x,1+h,1);"

(%i1) assume(h > 0 and c^2-1 < 0 and c > 0)
                                     2
(%o1)                       [h > 0, c  < 1, c > 0]
(%i2) integrate(-(c^2-(1-1/r))^((-1)/2),r,1+h,1)
                                   2                     2           2
                 2       sqrt(1 - c ) sqrt(h + 1) sqrt((c  - 1) h + c )
(%o2) (sqrt(1 - c ) atan(----------------------------------------------)
                                        2           2
                                      (c  - 1) h + c  - 1
     2                         2           2     4      2
 + (c  - 1) sqrt(h + 1) sqrt((c  - 1) h + c ))/(c  - 2 c  + 1)
                                 2
             2       c sqrt(1 - c )     3
   sqrt(1 - c ) atan(--------------) + c  - c
                          2
                         c  - 1
 - ------------------------------------------
                  4      2
                 c  - 2 c  + 1

You might want to try setting it up the same way in mathematica. It will certainly be simpler and easier to interpret after the change in variables.
Actually, there's no guarantee that this result comes out real just because there are no $i$'s in it. It does have one square root in it that looks like it could be imaginary. The way we know that it should be real is that it's a definite integral of a real integrand. If you then want to check whether it really is real, try just using some random numbers for $C$ and $h$. If the result comes out to be real, then that's not a coincidence. If it comes out complex, then one of the following may hold: (1) you've coded it incorrectly, (2) there is a bug in the CAS, or (3) there's an issue with branch cuts for the inverse tangent.
