# How can I compute the partial derivative under the line integral sign if the integrand is non-conservative?

Consider the following integral

$$I=\int_C^{\vec{y}_{0},\vec{y}} \vec{F}(\vec{x})\cdot d\vec{x}$$

where, for simplicity, $\vec{y}_{0}$ is some constant initial point on $C$ (which is an explicitly known curve) and $\vec{y}$ is some variable end point on $C$ and it is known that

$$\oint_{C_0} \vec{F}(\vec{x})\cdot d\vec{x} \neq 0$$

for any closed curve $C_{0}$.

Does it still make sense to speak of the gradient

$$\left(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial y_{2}},\frac{\partial}{\partial y_{3}}\right)I$$

of the integral with respect to $\vec{y} = \left(y_{1},y_{2},y_{3}\right)$ ?

If so, then three questions follow.

1) If this gradient exists then doesn't that contradict the premise that $\vec{F}$ is non-conservative?

2) Is there an equivalent of the Leibniz integral rule which specifies such a gradient?

3) How does it depend of the curve $C$?

Otherwise, how can we characterize the variational properties of this differentiation under the integral?

$I$, as you have given it, is not a well-defined function of $y$. There may be two different curves ending at $y$, yielding two different line integrals. Which one should I choose to compute the value of $I$ at any given $y$?
• The Curve in question is $C$ and has been given as part of the definition of $I$. So, there cannot be two curves as you say. There is exactly one curve and its name is $C$. – Mahlomola Daniel Cwele Nov 4 '17 at 15:34
• @MahlomolaDanielCwele If the curve $C$ is fixed, then yes, we get a well-defined function. But it is no longer a function of three independent variables. For example, if the curve is the $y_1$ axis, then how do we define $\partial I/\partial y_2$? This function does not appear to take any $y_2$ as input. – ziggurism Nov 4 '17 at 15:46
• The curve does not have to be one of the coordinate axes and, even if it is, the integral itself will still depend on the other coordinates simply because the integrand is assumed to depend of the global 3-D space. That is, $I=I(\vec{y},C)$ depends both on the chosen curve and on the coordinate system. We know that if the integrand was conservative, the curve $C$ would be irrelevant, resulting in $I=I(\vec{y})$. – Mahlomola Daniel Cwele Nov 4 '17 at 15:56
• To compute $\partial I/\partial y_2$, then we must compute $(I(y_2+h)-I(y_2))/h$. But $y_2+h$ is not the endpoint of your curve $C$. So either $C$ is not fixed in which case $I$ is not defined, or else the derivative is not defined. – ziggurism Nov 4 '17 at 15:58
• Those are not the only two options. The other option could be that $\frac{\partial I}{\partial y_{2}} = 0$ globally. That's what it means for a function to NOT depend on a given variable. But it does not imply that the function is not defined. – Mahlomola Daniel Cwele Nov 4 '17 at 16:03