Does $\int{\frac{1}{dx}}$ have any meaning in physics?

Question

Recently I came across this problem :

There are two identical parallel plates of length $L$ and breadth $B$ on the XZ plane . One plate passes through $Y = 0$ and the other passes through $Y = d$. Between them is a dielectric medium whose Dielectric constant changes as $\kappa = \kappa_0(3 + \frac{y}{L})$. Calculate the Capacity of the parallel plates.

The way to solve this problem is by considering a small portion of thickness $dy$ at a distance $y$ from the XZ plane.

Then we use the formula of $C = \frac{\kappa\epsilon_0 A}{d}$ to get the capacitance of the small portion. We denote this capacitance as $dC$

$$\begin{align} \Rightarrow dC &= \frac{\kappa \epsilon_0 LB}{dy} \\ \Rightarrow dC &= \frac{\kappa_0(3 + \frac{y}{L}) \epsilon_0 LB }{dy} \end{align}$$

We then take reciprocal on both sides and integrate it from $Y = 0$ to $Y = d$. The logic behind this is that since the layers of the dielectric are in series therefore we find the equivalent capacitance by integrating the reciprocals of capacitance of individual layers.

$$\Rightarrow \int\frac{1}{dC} = \int\frac{dy}{\kappa_0(3 + \frac{y}{L}) \epsilon_0 LB}\tag{i}$$

Solving the integral we get some value for $1/C$.

But the question is that this step might be logically correct but mathematically wrong because there is no meaning in $\int\frac{1}{dC}$ (or does it?). Can we really integrate undefined values by simply assigning some logical meaning behind it?

Answer 1

As a mathematical notation it is meaningless. But the overall derivation is correct, as you really pass to the continuous limit when switching from summing small elements $1/\Delta C$ to integration over $dy$. There are actually fine differences between an increment, a differential and a derivative, which textbooks often omit in this type of calculations.

A practical tip: write always $\Delta X$ for a small element and a summation sign $\Sigma$, and pass to $dx$ and integral $\int$ only at the last step of the derivation, when you actually take the number of the elements to infinity and pass to integration.

Answer 2

Your basic mistake is that you didn't do the differential correctly:

$dC= -\frac{3\kappa_0\epsilon_0 A}{y^2}dy$

You need to be careful when you integrate: because both sides of the integral diverge when $y$ goes to zero. The limits of integration for $C$ are $(3\kappa_0\epsilon_0 A/\epsilon, C_d)$, and those of $y$ are $(\epsilon, d)$. Once you do the integrals on both sides you can take the limit $\epsilon \rightarrow 0$ and get a finite result.

Answer 3

You can not write such an expression. Just consider the case were $dC$ and $dy$ are infinitesimal small. The left hand side of your expression is small, while the right hand side blows up. Therefore, this expression can not be true.

The correct way of doing such a thing is to write the capacitance as a function of the distance,$C = C(y)$, and take the derivative $$ \frac{dC}{dy} = ... $$ Finally, you multiply both sides with $dy$. Although the last step is mathematically questionable, you end up with the correct expression for $dC$. Finally consider the case of "many" capacitance in series.