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I have been posed the following question, which I am unable to answer:

Let $a_1,a_2\in\mathcal{C}^{\infty}(\Bbb{R}^2,\Bbb{R})$ be infinitely differentiable functions such that $\frac{\partial a_1}{\partial x_2}=\frac{\partial a_2}{\partial x_1}$, and define $$f(x,y):=x_1\cdot\int_0^1a_1(tx_1,tx_2)\ dt+x_2\cdot\int_0^1a_2(tx_1,tx_2)\ dt.$$ Show that $\tfrac{\partial f}{\partial x_i}=a_i$ for $i\in\{1,2\}$.

Unfortunately I lack even the most basic knowledge of calculus. Here's my attempt at a solution:

First by linearity of differentiation and second by the product rule and then cleaning up we find \begin{align*} \frac{\partial f}{\partial x_1}(x_1,x_2) &=\frac{\partial}{\partial x_1}\left(x_1\cdot\int_0^1a_1(tx_1,tx_2)\ dt\right)+\frac{\partial}{\partial x_1}\left(x_2\cdot\int_0^1a_2(tx_1,tx_2)\ dt\right)\\ &=x_1\cdot\frac{\partial}{\partial x_1}\left(\int_0^1a_1(tx_1,tx_2)\ dt\right)+\int_0^1a_1(tx_1,tx_2)\ dt\cdot\frac{\partial x_1}{\partial x_1}\\ &\hspace{10pt}+x_2\cdot\frac{\partial}{\partial x_1}\left(\int_0^1a_2(tx_1,tx_2)\ dt\right)+\int_0^1a_2(tx_1,tx_2)\ dt\cdot\frac{\partial x_2}{\partial x_1}\\ &=\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\frac{\partial}{\partial x_1}\int_0^1a_1(tx_1,tx_2)\ dt+x_2\cdot\frac{\partial}{\partial x_1}\int_0^1a_2(tx_1,tx_2)\ dt. \end{align*} Now changing the order of integration and differentiation, which is allowed for whatever reason, and subsequently applying the chain rule and substituting $\frac{\partial a_1}{\partial x_2}=\frac{\partial a_2}{\partial x_1}$ we find that \begin{align*} \frac{\partial f}{\partial x_1}(x_1,x_2) &=\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\int_0^1\frac{\partial a_1(tx_1,tx_2)}{\partial x_1}\ dt+x_2\cdot\int_0^1\frac{\partial a_2(tx_1,tx_2)}{\partial x_1}\ dt\\ &=\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\int_0^1\frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\cdot t\ dt+x_2\cdot\int_0^1\frac{\partial a_2}{\partial x_1}(tx_1,tx_2)\cdot t\ dt\\ &=\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\int_0^1\frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\cdot t\ dt+x_2\cdot\int_0^1\frac{\partial a_1}{\partial x_2}(tx_1,tx_2)\cdot t\ dt. \end{align*} These integrals look good for integration by parts with respect to $t$ and $\tfrac{\partial a_i}{\partial x_1}(tx_1,tx_2)$, which yields \begin{align*} \frac{\partial f}{\partial x_1}(x_1,x_2) &=\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\left(\left[t\cdot\int\frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\ dt\right]_0^1-\int_0^1\int\frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\ dt\ dt\right)\\ &\hspace{10pt}+x_2\cdot\left(\left[t\cdot\int\frac{\partial a_2}{\partial x_1}(tx_1,tx_2)\ dt\right]_0^1-\int_0^1\int\frac{\partial a_2}{\partial x_1}(tx_1,tx_2)\ dt\ dt\right).\\ \end{align*} This is where my bright ideas end; I've juggled the terms around a bit but I see no way of evaluating any of these expressions. Any help would be much appreciated.

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Up to

$$\frac{\partial f}{\partial x_1} (x_1,x_2) =\int_0^1a_1(tx_1,tx_2)\ dt+x_1\cdot\int_0^1\frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\cdot t\ dt+x_2\cdot\int_0^1\frac{\partial a_1}{\partial x_2}(tx_1,tx_2)\cdot t\ dt,$$

it is good. But then you took the wrong turn. The trick is that the sum of the integrands is a derivative with respect to $t$. Write

$$\begin{align} \frac{\partial f}{\partial x_1} (x_1,x_2) &= \int_0^1a_1(tx_1,tx_2)\,dt + \int_0^1 t\cdot\left( \frac{\partial a_1}{\partial x_1}(tx_1,tx_2)\cdot x_1 + \frac{\partial a_1}{\partial x_2}(tx_1,tx_2)\cdot x_2\right)\,dt\\ &= \int_0^1a_1(tx_1,tx_2)\ dt + \int_0^1 t\cdot \frac{\partial}{\partial t}a_1(tx_1,tx_2)\,dt\\ &= \int_0^1 \frac{\partial}{\partial t} \left(t\cdot a_1(tx_1,tx_2)\right)\,dt\\ &= 1\cdot a_1(1\cdot x_1,1\cdot x_2) - 0 \cdot a_1(0,0)\\ &=a_1(x_1,x_2). \end{align}$$

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  • $\begingroup$ What a clever way to rewrite things, thank you very much! $\endgroup$ – user125421 Feb 5 '14 at 0:00

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