Asking about a step regarding indices in deriving the Curvature tensor from the geodesic equation.
Starting from
$$ \frac{d v^a}{du} = - \Gamma^a_{bc}v^b \frac{dx^c}{du}$$
we integrate
$$v^a(u) = v^a(p) - \int_p^u \Gamma^a_{bc} v^b(u) \frac{dx^c}{du}du$$
then Taylor expand $\Gamma^a_{bc}(u)$ as
$$\Gamma^a_{bc}(u) = \Gamma^a_{bc}(p) + \partial_d \Gamma^a_{bc}(p)(x^d(u)-x^d(p))$$
to find
\begin{align} v^a(u) &= v^a(p) - \int_p^u [\Gamma^a_{bc}(p) + \partial_d \Gamma^a_{bc}(p)(x^d(u)-x^d(p))]v^b(u) \frac{dx^c}{du}du \\ &= v^a(p) - \int_p^u \Gamma^a_{bc}(p) v^b(u) \frac{dx^c}{du}du - \int_p^u \partial_d \Gamma^a_{bc}(p)(x^d(u)-x^d(p))]v^b(u) \frac{dx^c}{du}du \\ &= v^a(p) - \Gamma^a_{bc}(p) \int_p^u v^b(u) \frac{dx^c}{du}du - \partial_d \Gamma^a_{bc}(p) \int_p^u (x^d(u)-x^d(p))]v^b(u) \frac{dx^c}{du}du \end{align}
Now I'm supposed to expand the $v^b(u)$ term using a formula like
$$v^a(u) = v^a(p) - \Gamma^a_{bc}(p) v^b (p)[x^c(u) - x^c(p)]$$
to derive
$$v^a(u) = v^a(p) - \Gamma^a_{bc}(p) v^b(p) \int_p^u \frac{dx^c}{du}du - (\partial_d \Gamma^a_{bc} - \Gamma^a_{ec} \Gamma^e_{bd})v^b(p) \int_p^u (x^d(u) - x^d(p)) \frac{dx^c}{du}du$$
but it doesn't work when I try, I don't see where the $e$ terms on both $\Gamma$'s come from or how to change the formula for $v^a(u)$ I'm supposed to use.
What do I do?
You need to reindex the formula you're trying to use, both so that it gives you $v^b(u)$, and so that its dummy summation indices are different from the ones you're already using:
$$ v^b(u) = v^b(p) - \Gamma^b_{ed}(p) v^e (p)[x^d(u) - x^d(p)]\;. $$
Substituting this for $v^b(u)$ yields the first integral term in the result, plus a term
$$ \Gamma^a_{bc}(p)\Gamma^b_{ed}(p) v^e (p) \int_p^u [x^d(u) - x^d(p)] \frac{\mathrm dx^c}{\mathrm du}\mathrm du\;. $$
Now you just need to swap the dummy summation indices $b$ and $e$ to make this coincide with the last term in the result.