Let $\alpha:[a,b] \rightarrow \mathbb{R}^n$ be a path and let $\alpha^{-1}$ denote the inverse path given by $\alpha^{-1}(t)=\alpha(a + b - t)$ Also, let $F$ be a vector field on $\mathbb{R}^n$ and assume appropriate hyphotheses regarding smoothness, domains, etc. I am trying to show that
$ \int\limits_{\alpha^{-1}} F \cdot d(\alpha^{-1}) = -\int\limits_{\alpha} F \cdot d\alpha $
This is an intuitive result but, technically, I can't seem to manage to get the signs right and I'm sure I'm making a silly mistake. Here are my computations:
$ \int\limits_{\alpha^{-1}} F \cdot d(\alpha^{-1}) = $
$ =\int\limits_{a}^{b}F(\alpha^{-1}(t)) \cdot [\alpha^{-1}(t)]^\prime dt $
$ =\int\limits_{a}^{b}F(\alpha(a + b -t)) \cdot \alpha^{\prime}(a + b - t)dt $
Now, let $u = a + b - t$. Then, $t = a \implies u = b$ and $t = b \implies u = a$. Furthermore, $dt = -du$. Therefore, we can rewrite the last integral as
$ - \int\limits_{b}^{a}F( \alpha(u)) \cdot \alpha^{\prime}(u)du $
$ = \int\limits_{a}^{b}F( \alpha(u)) \cdot \alpha^{\prime}(u)du $
$ = \int\limits_{\alpha} F \cdot d\alpha $
Which is, obviously, not what I'm trying to show. There are two sign changes here which cancel one another out. The first is due to the fact that $dt = -du$ and the second because I am reversing the order the integration limits so that the integral fits the definition of the path integral.
So, my question is, what step is in error?