Differentiation Under Integral Sign

Theorem

Let $f \left({x, y}\right)$ and $\dfrac{\partial f}{\partial x} \left({x, y}\right)$ be continuous functions of $x$ and $y$ on $D = \left[{{x_1} \, . \, . \, {x_2}}\right] \times \left[{{a} \, . \, . \, {b}}\right]$.

Then:

$\displaystyle \frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y$

for $x \in \left[{{x_1} \, . \, . \, {x_2}}\right]$.

Proof

Define $\displaystyle G \left({x}\right) = \int_a^b f \left({x, y}\right) \mathrm dy$.

The continuity of $f$ ensures that $G$ exists.

Then by linearity of the integral:

$\displaystyle \frac{\Delta G}{\Delta x} = \frac{G \left({x + \Delta x}\right) - G \left({x}\right)} {\Delta x} = \int_a^b \frac{f \left({x + \Delta x, y}\right) - f \left({x, y}\right)} {\Delta x}\mathrm dy$

We want to find the limit of this quantity as $\Delta x$ approaches zero.

For each $y \in \left[{{a} \, . \, . \, {b}}\right]$, we can consider $f_y \left({x}\right) = f \left({x, y}\right)$ as a separate function of the single variable $x$, with $\displaystyle \frac{\mathrm df_y}{\mathrm dx} = \frac{\partial f}{\partial x}$.

Thus by the Mean Value Theorem, there is a number $c_y \in \left({x, x + \Delta x}\right)$ such that $\displaystyle f_y \left({x + \Delta x}\right) - f_y \left({x}\right) = \frac {\mathrm df_y} {\mathrm dx} \left({c_y}\right) \Delta x$.

That is:

$\displaystyle f(x + \Delta x, y) - f(x, y) = \frac{\partial f}{\partial x}(c_y, y)\Delta x$

Therefore:

$\displaystyle \frac {\Delta G} {\Delta x} = \int_a^b \frac {\partial f} {\partial x} \left({c_y, y}\right)\mathrm dy$

Now, pick any $\epsilon > 0\ $ and set $\epsilon_0 = \dfrac{\epsilon}{b - a}$.

Since $\dfrac{\partial f}{\partial x}$ is continuous on the compact set $D$, it is uniformly continuous on $D$.

Hence for each $x$ and $y$:

$\displaystyle \left\vert{\frac {\partial f} {\partial x} \left({x + h, y}\right) - \frac {\partial f} {\partial x} \left({x, y}\right)}\right\vert < \epsilon_0$

whenever $h$ is sufficiently small.

And since $x < c_y < x + \Delta x\ $, it follows that for sufficiently small $\Delta x\ $ that:

$\displaystyle \left\vert{\frac {\partial f} {\partial x} \left({c_y, y}\right) - \frac {\partial f} {\partial x} \left({x, y}\right)}\right\vert < \epsilon_0$

regardless of our choice of $y$.

So we can say:

	\(\displaystyle \)	\(\displaystyle \left\vert{\lim_{\Delta x \to 0} \frac{\Delta G}{\Delta x} - \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm dy}\right\vert\)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} \left\vert{\int_a^b \frac{\partial f}{\partial x} \left({c_y, y}\right) - \frac{\partial f}{\partial x} \mathrm dy}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \lim_{\Delta x \to 0} \int_a^b \left\vert{\frac{\partial f}{\partial x} \left({c_y, y}\right) - \frac{\partial f}{\partial x} \left({x, y}\right)}\right\vert \mathrm dy\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \int_a^b \epsilon_0 \mathrm dy\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \epsilon\)	\(\displaystyle \)

But since $\epsilon$ was arbitrary, it follows that $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta G}{\Delta x} = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right)\mathrm dy$ and the theorem is proved.

$\blacksquare$