Why is it necessary to represent a function as $h(x(t),y(t))$ before integrating?

Question

I was reading a proof on the following:

Two continuous complex-valued functions $P,Q,$ we have that $\int Pdx+Qdy$ is independent of path if and only if $Pdx+Qdy$ is exact.

In the proof, we went through a procedure to get the following:

Suppose $h$ is analytic such that $dh=Pdx+Qdy.$ For any path $\gamma$ from $A$ to $B,$ we have that $$\int_{\gamma}dh=\int_\gamma dh(x(t),y(t))=h(B)-h(A).$$

The proof of how this was achieved made sense to me, but what I did not understand was what we achieved from doing this procedure. Why did we need to show that $\int_{\gamma}dh=\int_\gamma dh(x(t),y(t))$ before integrating? And why does $\int_\gamma dh(x(t),y(t))=h(B)-h(A)?$

Any help will be much appreciated!

Answer 1

The curve $\gamma$ is expressed as a parameterization $\gamma(t) = \big(x(t),y(t)\big) \;.$ This means that the integral $$\int_\gamma dh = \int_\gamma P(x,y)dx + \int_\gamma Q(x,y)dy =\int_{\gamma(t)} P(x(t),y(t))dx + \int_{\gamma(t)} Q(x(t),y(t))dy $$ can undergo the change of variables $dx = \frac{dx}{dt}dt$ and $dy = \frac{dy}{dt}dt$ to achieve

$$\int_{\gamma(t)} P(x(t),y(t))\frac{dx}{dt}dt + \int_{\gamma(t)} Q(x(t),y(t))\frac{dy}{dt}dt = \int_{\gamma(t)} \bigg(P(x(t),y(t))\frac{dx}{dt} + Q(x(t),y(t))\frac{dy}{dt}\bigg) \; dt \; .$$

Since we are integrating over $t,$ we are ensured to be integrating over the curve provided with no deviation from it. In other words, this is how you integrate a differential one-form $dh$ over a curve $\gamma(t).$ You must parameterize it.

The equation you provided $$\int_\gamma dh(x(t),y(t))=h(B)-h(A)$$

directly implies that integration over the one-form $dh$ is pathway independent. In other words, no matter what curve $\gamma$ we choose with start point $A$ and end point $B,\; \;$ $\int_\gamma dh$ will be the same and thus only depend on the end points $A,B.$

Answer 2

The reason is that you need to somehow describe the path, which is an open curve. You could do this with an implicit equation $f(x,y)=0$, and some additional trick to delimit the endpoints, but this is not convenient. A parametric representation $x=f(t),y=g(t)$ is more handy. With that, the path integral turns to an ordinary integral on a scalar variable, $t$.