D'Inverno derivation of Schwarzschild solution

[PhyPsy]

If you happen to have D'Inverno's Introducing Einstein's Relativity, this is on page 187. He has reduced the metric to non-zero components:
g_{00}= e^{h(t)}(1-2m/r)
g_{11}=-(1-2m/r)^{-1}
g_{22}=-r^2
g_{33}=-r^2\sin^2\theta
The final step is a time coordinate transformation that reduces g_{00} to 1-2m/r. This is achieved by making e^{h(t')}=1, so h(t')=0. He does this with the relation
t'=\int^t_c e^{\frac{1}{2}h(u)}du, c is an arbitrary constant
I suppose that, since c is arbitrary, I can assign whatever value to c to make h(t')=0, but why use this particular integral as the relation between t and t'? Is there something special about this integral?

 

[Bill_K]

This is pretty easy. You want (1 - 2m/r) dt'² = e^h(t)(1 - 2m/r) dt², so take dt' = e^h(t)/2 dt

 

[PhyPsy]

Ah, OK...but why would the relation not be t'=\int e^{h(t)/2}dt? Instead, they have it as a definite integral from c to t.