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 $dg=\alpha$?

  • $\begingroup$ You only need to use Leibniz's integral rule 1 time and not 4 times. Check out my comment to this question. It's all about understanding what a single integral of a multivariable function is. Once you see this, then use what we know from single variable calculus: $\frac{d}{dx}\int_a^x f = f(x)$. Therefore your first line, 3rd line, and 4th line under the link you give is just applying this result. Only the 2nd line do you need leibniz $\endgroup$
    – DWade64
    Commented Oct 3, 2018 at 19:25

1 Answer 1


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.

  • $\begingroup$ thank you @Ted Shifrin $\endgroup$
    – Desunkid
    Commented Oct 4, 2018 at 13:07
  • $\begingroup$ You're welcome. Please accept the answer when you are satisfied and have no further questions, so that the question will not stay on the "unanswered" list. $\endgroup$ Commented Oct 4, 2018 at 15:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .