# How to prove the integral formulae of the inverse path $\alpha^-$ and the product path $\alpha\beta$?

I need help with this problem:

Let $$f:S\subset\mathbb{R}^n\rightarrow\mathbb{R}$$ be continuous on $$S$$, and let $$\alpha:[a,b]\subset\mathbb{R}\rightarrow\mathbb{R}^n$$ and $$\beta:[c,d]\subset\mathbb{R}\rightarrow\mathbb{R}^n$$ be piecewise $$C^1$$ (continuously differentiable) paths in $$S$$, such that $$\alpha(b)=\beta(c)$$. Prove the following integral formulae concerning the inverse path $$\alpha^-$$ and the product path $$\alpha\beta$$: $$\int_{\alpha^-}f\ ds=\int_{\alpha}f\ ds$$ $$\int_{\alpha\beta} f \ ds=\int_{\alpha}f \ ds\ + \int_{\beta}f \ ds$$

I don't know if I'm correct, but I think that $$\alpha^-(t)=\alpha(-t)$$, so $$\alpha^-:[-b,-a]\rightarrow\mathbb{R}^n$$. Using this I tried to rpove the first one like this: $$\int_{\alpha^-}f\ ds=\int_{-b}^{-a} f(\alpha^-(t))\Vert \alpha{^-}'(t)\Vert dt=\int_{-b}^{-a} f(\alpha(-t))\vert(-1)\vert\Vert\alpha(-t)\Vert dt$$ let $$u=-t$$, thus $$du=-dt$$ $$-\int_a^b-f(\alpha(u))\Vert\alpha'(u)\Vert du=\int_a^bf(\alpha(u))\Vert\alpha'(u)\Vert dt=\int_\alpha f\ ds$$ Am I correct? For the second one, I don't know how to prove it. I think that $$\alpha\beta$$ woudl be the path that goess from $$a$$ to $$d$$, right?

The first one looks correct, for the second, you have to join the paths $$\alpha, \beta$$ into one parameterisation $$\gamma$$. One way to do this is by letting,

$$\gamma : [a, b + d - c] \to \mathbb{R}^n$$ with,

$$\gamma(t) = \begin{cases}\alpha(t), \ t \in [a,b] \\ \beta(t - b + c), \ t \in [b, b + d- c] \end{cases}$$

See here how I have combined the two paths into one interval from $$a$$ to $$b + d -c$$. I have not changed the paths at all here, so that $$\gamma = \alpha \beta$$. That means,

$$\int_\gamma f \ \mathrm{d}s = \int_a^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \int_a^b f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t + \int_b^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \dots$$

I'll let you finish the proof, at this point you have to use the definition of $$\gamma$$. You may also want to verify that $$\gamma$$ is in fact $$C^1$$ at every point except at $$t = b$$, which won't affect the integral since we only require $$\gamma$$ be piecewise $$C^1$$.

• I don't understand why the interval goes from $a$ to $b+d-c$. Also, why $gamma$ is not $C^1$ at $t=b$? – davidllerenav Apr 20 '19 at 3:22