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We have that $f : \mathbb{R}^2 \mapsto \mathbb{R}, f \in C^2$ and $h= \nabla f = \left(\frac{\partial f}{\partial x_1 },\frac{\partial f}{\partial x_2 } \right)$, $x=(x_1,x_2)$.

Now the proposition I try to show says that

$$\int_0^1 \! \langle \nabla f(x \cdot t),x \rangle\,dt \,= \int_0^{x_1} \! h_1(t,0) ,dt \, +\int_0^{x_2} \! h_2(x_1,t) \,dt \,$$

I know that $\langle \nabla f(x \cdot t),x \rangle=d f(tx) \cdot x$ but it doesn't seem to help, maybe you have to make a clever substitution? (Because the range of integration changes). Thanks in advance.

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    Is $f$ really $\mathbb{R}^2 \to \mathbb{R}^2$, or $f:\mathbb{R}^2 \to \mathbb{R}$?2011-05-17
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    You are right I am sorry for this severe mistake. I looked it up and the notation was misleading but it doesn't seem to hold for $\mathbb{R}^2$.2011-05-17

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Before you read this answer, fetch a piece of paper and draw the following three points on it: $(0,0)$, $(x_{1},0)$ and $(x_{1},x_{2})$. These are the corners of a right-angled triangle whose hypotenuse I'd like to call $\gamma$ whose side on the $x_1$-axis I call $\gamma_{1}$ and whose parallel to the $x_2$-axis I call $\gamma_2$.

More formally, let $\gamma: [0,1] \to \mathbb{R}^2$ be the path $t \mapsto tx$. Similarly, let $\gamma_{1} : [0,1] \to \mathbb{R}^2$ be the path $t \mapsto (tx_{1},0)$ and $\gamma_{2} : [0,1] \to \mathbb{R}^2$ be the path $t \mapsto (x_{1}, tx_{2})$.

The integral on the left hand side can be written as $$\int_{0}^{1} df(\gamma(t))\cdot\dot{\gamma}(t)\,dt = \int_{0}^{1} \frac{d}{dt}(f \circ \gamma)(t)\,dt = f(\gamma(1)) - f(\gamma(0)) = f(x_1, x_2) - f(0,0).$$

Similarly, after some simple manipulations the right hand side is equal to $$\int_{0}^{1} \frac{d}{dt} (f \circ \gamma_{1})(t)\,dt + \int_{0}^{1} \frac{d}{dt} (f \circ \gamma_2)(t)\,dt = \left( f(\gamma_{1}(1)) - f(\gamma_{1}(0))\right) + \left(f(\gamma_2 (1)) - f(\gamma_2(0))\right)$$ and as $\gamma_1 (1) = (x_1,0) = \gamma_2 (0)$ two terms cancel out and what remains is $f(x_{1},x_{2}) - f(0,0)$.

Thus the left hand side and the right hand side are equal.

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    Thank you for figuring this out, I also like the geometric interpretation of the equality a lot.2011-05-17
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    @user: Oh, that was quick! Well, what is really going on is that the integral of $df(\gamma(t))\cdot\dot{\gamma}(t)$ only depends on the end points of the path $\gamma$. On the left hand side we're going straight from the origin to $(x_1,x_2)$ and on the right hand side we make a detour via $(x_1,0)$.2011-05-17
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    Yes I think the main point was that you need to use the Fundamental theorem of calculus to simplify the integral. And the chain rule backwards, which is also very clever.2011-05-17
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    @user: Yes, that's it exactly. In a sense I figured out a substitution. But I couldn't have found it without recognizing the terms $\langle \nabla f \circ \gamma, \dot \gamma \rangle$ and similarly with $\gamma_{1}$ and $\gamma_2$ in place of $\gamma$. In the first integral on the right hand side I only have $h_{1}$, so $\dot{\gamma_1} = (1,0)$ and in the second one I only have $h_2$, so $\dot{\gamma_2} = (0,1)$. So I looked for paths that fit together. Do you see what I mean? (thanks by the way, but this is not cleverness, just a bit of practice :)2011-05-17
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    Yes I see it, the approach is also tricky, I wouldn't have come up with the idea to rewrite the $x$ as a derivate of $tx$. I feel with the right definitions the proof is quite easy to follow (and elegant) but hard to come up with.2011-05-17
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    @user: You can also do it with [Green's theorem](http://en.wikipedia.org/wiki/Green's_theorem) by taking everything on the right hand side and walking through $\gamma$ backwards to get rid of the sign, i.e. $t \mapsto (1-t)x$. If you don't know Green's thm just look at the first equation in the first formula on the page I'm linking to: $C$ is the boundary of the triangle (that is, the three paths), $D$ is the triangle itself (the region in the plane), and $(L,M) = h$. Since $f$ is $C^2$ we have $\frac{\partial L}{\partial x_2} = \frac{\partial M}{\partial x_1}$ since $f$ is $C^2$. ....2011-05-17
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    ... Then all comes down to $\nabla \times \nabla f = 0$. I'm sure you'll learn that soon in case I'm not making much sense now.2011-05-17
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    That theorem is too advanced for me I only know single dimensional integrals but I see on the image that it does what I need :-)2011-05-18
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    @user: I guessed that it might be too advanced and it sounds like you got my point: I directed you to it so you know about it, as it's a really cool and useful result. One last thing: the formula you asked about can be used as one of the key insights towards a proof of Green's theorem, so it's not just a curiosity. See you around!2011-05-18
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    I thought I got it all but I think there is some final step missing to bringt it to the needed form. I tried to expand the second term to $\int_{0}^{1} \frac{d}{dt} (f \circ \gamma_{1})(t)\,dt + \int_{0}^{1} \frac{d}{dt} (f \circ \gamma_2)(t)\,dt=\int_{0}^{1} x_1 \cdot \frac{d}{dt} f_1(t x_1,0)\,dt + \int_{0}^{1} x_2 \cdot \frac{d}{dt} f_2(x_1,t x_2)\,dt$ but even the range of integration still seems wrong (how do I get the x_1 instead of the 1) and what happens to the inner arguments?2011-05-18
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    @user: If you substitute $u = u(t) = x_{1}t$ then $du = x_{1}\,dt$ and $$\int_{0}^{1} h_{1}(x_{1} t,0) \cdot x_{1}\,dt = \int_{0}^{x_1} h_{1}(u,0)\,du,$$ right? That's what I meant with "After some simple manipulations...". I think you can handle the second integral on your own.2011-05-18
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    Sorry for not seeing things that are obvious to you. The problem was I always thought the variable of integration is the same because both were named $t$. Thank you, now it is completely clear.2011-05-18
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    @user: Hey, no need to apologize! This is a perfectly legitimate question and I didn't mean to imply that it should be immediately obvious. You're always welcome to ask. That's what this site is here for, after all... :)2011-05-18
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    @user: In fact, if you prefer, you can take the paths $\tilde{\gamma}_1 : [0,x_1] \to \mathbb{R}^2, t \mapsto (t,0)$ and $\tilde{\gamma}_2 : [0,x_{2}] \to \mathbb{R}^2, t \mapsto (x_1, t)$ in place of $\gamma_1$ and $\gamma_2$ in my answer. Then there's no need for the somewhat unnecessary substitution step. But I prefer to have all my paths defined on $[0,1]$.2011-05-18
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    I see but I also prefer your initial solution. It is easier to see whats going on there.2011-05-18