June 4, 2021

Leibniz’s rule (LR-1) states:

Let $f(x, \beta)$ be continuous and have a continuous derivative $\frac{\partial}{\partial \beta}$ in a domain of $x\beta-$ plane that includes the rectangle $a \le x \le b, \beta_1 \le \beta \le \beta_2,$

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx =\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

I will derive LR-1 semi-rigorously as follows:

Let

$g(t) = \int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1-1)$

Integrate (1-1) with respect to $t$ from a constant $\alpha$ to a variable $\beta$ , we have

$\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{\alpha}^{\beta}\left(\int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx\right)\;dt$

$\overset{(\star)}{=}\int\limits_{a}^{b}\left(\int\limits_{\alpha}^{\beta} \frac{\partial}{\partial t}f(x, t)\;dt\right)\;dx$

$=\int\limits_{a}^{b}f(x, \beta) - f(x, \alpha)\; dx$

$=\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.$

That is,

$\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.\quad\quad\quad(1-2)$

While $\int\limits_{\alpha}^{\beta}g(t)\;dt$ and $\int\limits_{a}^{b}f(x,\beta)\;dx$ are functions of $\beta$ , $\int\limits_{a}^{b}f(x,\alpha)\;dx$ is a constant.

Since

$\frac{d}{d\beta}\int\limits_{\alpha}^{\beta} g(t)\;dt \overset{\textbf{FTC}}{= }g(\beta), \quad\frac{d}{d\beta}\left(\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx\right)=\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx,$

differentiate (1-2) with respect to $\beta$ gives

$g(\beta) = \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx\overset{(1-1)}{\implies} \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx= \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx;$

i.e.,

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx = \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

In the three-dimensional $x, y, z$ -space, the double integral of a continuous function with two independent variables, $V=\iint_{R} f(x, y) dx dy$ , may be interpreted as a volume between the surface $z=f(x, y)$ and the $x, y$ -plane: