Consider two functions $P,Q : \mathbb{R}^2 \to \mathbb{R}$ related by

$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} $$

We ask the question, are $P,Q$ derivable from some other function $f$ where

$$P = \frac{\partial f}{\partial x}\;\;\;\; \\ Q = \frac{\partial f}{\partial y}\;\; ? $$

The answer is yes and the proof with (definite) integrals is the following. If the last two equations are true statements, then taking the single integral with respect to $y$ gives

$$ f = \int_{y_0}^y Q(x,t) dt + R(x)$$

And finally to fit in the last requirement $P = \partial f/\partial x$, we require that $$\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}\int_{y_0}^y Q(x,t)dt + \frac{dR}{dx} = P(x,y)$$

Using Leibniz's integral rule to interchange operations, we get

$$ \int_{y_0}^y \frac{\partial Q}{\partial x}(x,t)dt + \frac{dR}{dx} = P(x,y)$$

$$ \int_{y_0}^y \frac{\partial P}{\partial y}(x,t)dt + \frac{dR}{dx} = P(x,y)$$

$$ P(x,y) - P(x,y_0) + R'(x) = P(x,y)$$

$$ R'(x) = P(x,y_0)$$ $$ R = \int_{x_0}^x P(t,y_0) dx + C$$

In total, $f$ is of the form

$$ f(x,y) = \int_{x_0}^x P(t,y_0) dt + \int_{y_0}^y Q(x,t) dt + C$$

How do I get to this result with indefinite integrals? Is it possible? I start out the proof the same way:

$$ f = \int Q \; dy + R(x)$$


$$\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}\int Q \; dy + R'(x) = P $$

Assuming I can still use Leibniz's integral rule?

$$ \int \frac{\partial Q}{\partial x} \; dy + R'(x) = P$$

$$ \int \frac{\partial P}{\partial y} \; dy + R'(x) = P$$


$$P + S(x) + R'(x) = P ? $$ $$ R'(x) = S(x)? $$

$$R = \int S(x)\; dx + C $$

So $f$ must take the form $$ f(x,y) = \int Q \; dy + \int S(x) \; dx + C$$

However I don't see how this result is the same as the previous one. Or rather, it does seem to be the same if

$$ \int S(x)\;dx = \int_{x_0}^x P(t,y_0)dt$$

Did I go wrong when using indefinite integrals? How do you prove the form of $f$ using indefinite integrals? Thank you in advance for any help


1 Answer 1


The error seems to be that I applied Leibniz's integral rule when I shouldn't have. The difference between $\int f$ and $\int_a^x f$ is that the first is an arbitrary antiderivative while the second is a particular antiderivative. Because of this, I assumed that I could use Leibniz's integral rule on an arbitrary antiderivative (I assumed "they are both antiderivatives so why not?")

However, it seems like I can't do this (which I'm still not sure why, from an intuitive standpoint). I should only apply Leibniz's integral rule to definite integrals.

  • $\begingroup$ It may have something to do with the fact that $f$ is a function of 2 variables. If $f$ were a single variable function, then I think $d/dx \int f = \int d/dx f$. However Leibniz's integral rule, as stated, requires a full derivative outside the integral sign (after integration the 2 variable function becomes a 1 variable function). However If I just do $\int f(x,y) dy$, this is still a 2 variable function, requiring a partial derivative $\endgroup$
    – DWade64
    Oct 7, 2018 at 12:49

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