A Semi-Rigorous Derivation of Leibniz’s Rule

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:

Fig. 1 V = \iint_R f(x,y) dx dy

We see from Fig. 1 that on one hand,

V=\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy,\quad\quad\quad(2-1)

but on the other hand,

V=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(2-2)

Since (2-1) and (2-2) amounts to the same thing, it must be true that

\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(\star)

In other words, the order of integration can be interchanged.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s