Let $\alpha= f_1(x_1,x_2)dx_1 + f_2(x_1,x_2)dx_2$ such that
$d\alpha=0$. Define a function $g$ by $$g(x_1,x_2)= \int_{0}^{x_1} f_1(t,x_2) dt + \int_{0}^{x_2} f_2(0,t)dt$$ Show that $dg=\alpha$
By $d\alpha=0$ we get $\dfrac{\partial{f_2}}{\partial{x_1}}-\dfrac{\partial{f_1}}{\partial{x_2}}=0$ and
$$dg= \dfrac{\partial}{\partial{x_1}} ( \int_{0}^{x_1} f_1(t,x_2) dt ) dx_1+\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_1} f_1(t,x_2) dt) dx_2$$
$$+\dfrac{\partial}{\partial{x_1}}(\int_{0}^{x_2} f_2(0,t) dt) dx_1 +\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_2} f_2(0,t) dt) dx_2$$
By Leibniz integral rule
$\dfrac{\partial}{\partial{x_1}} ( \int_{0}^{x_1} f_1(t,x_2) dt ) =f_1(x_1,x_2) + ( \int_{0}^{x_1}\dfrac{\partial}{\partial{x_1}} f_1(t,x_2) dt ) $
$\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_1} f_1(t,x_2) dt)=(\int_{0}^{x_1} \dfrac{\partial}{\partial{x_2}}f_1(t,x_2) dt)$
$\dfrac{\partial}{\partial{x_1}}(\int_{0}^{x_2} f_2(0,t) dt)=(\int_{0}^{x_2} \dfrac{\partial}{\partial{x_1}}f_2(0,t) dt)$
$\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_2} f_2(0,t) dt)=f_2(0,x_2)+(\int_{0}^{x_2}\dfrac{\partial}{\partial{x_2}} f_2(0,t) dt)$
Do i have any mistake? And can you continue to compute those and indicate $dg=\alpha$?
Best Answer
This is fine so far. Now $$ \frac{\partial}{\partial x_1} f_1(t,x_2) = \frac{\partial}{\partial x_1} f_2(0,t) = \frac{\partial}{\partial x_2} f_2(0,t) = 0.$$ Why? All that you have left to do is use the hypothesis that $d\alpha=0$ to rewrite $$\int_0^{x_1} \frac{\partial}{\partial x_2} f_1(t,x_2)dt = \int_0^{x_1}\frac{\partial}{\partial x_1} f_2(t,x_2)dt,$$ and finish the computation.