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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?

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up vote 3 down vote accepted

The mistake is in the derivative of $\alpha^{-1}$. That is from step 2 to 3 in your reasoning. It should be $-\alpha'(a+b-t)$.

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But, from step 2 to step 3 I'm not actually evaluating the derivative; I'm just replacing $\alpha^{-1}$ with $\alpha(a + b - t)$. That is, $\frac{d}{dt}( \alpha^{-1}(t) ) = \frac{d}{dt}(\alpha(a + b - t))$ – ItsNotObvious Jun 18 '11 at 22:27
Yes, but you have to apply the chain rule. The derivative of $\alpha(a+b-t)$ is that of $\alpha$ evaluated in $a+b-t$ times the derivative of $a+b-t$ with respect to $t$. – Raskolnikov Jun 19 '11 at 8:43
OK, I think I understand the situation after thinking about it a little more. Basically, the derivative of $\alpha$ must actually taken in order for the following substitution step to actually make sense. Thanks. – ItsNotObvious Jun 19 '11 at 13:49

I think you forgot to apply the chain rule when taking the derivative of $\alpha(a+b-t)$. There should be an additional factor $-1$ from the $-t$ term.

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If I remember well , using c for the curve, you can refer to the original integrand as f(c(a+t(b-a)))c'(a+t(b-a)), since this parametrization agrees with the standard ordering of the real line, i.e., the parametrization a+t(b-a) will output the points in the order you want them. Then the reverse orientation is the reverse ordering-- b-t(a-b)--but the image f(c(b-t(a-b)) is still the same as the original; f(x) is the same, only seen from a different direction. Note that your choices of parametrization for the curves c and $c^-1$ do not cancel out to the identity (giving you the starting point). Then, by the chain rule , c'(b-t(b-a)) will give you the negative sign.

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