Definite Integral of Partial Derivative

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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:

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

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


Proof 1

From Leibniz Integral Rule:

$\displaystyle \frac \d {\d y} \int_{a \left({y}\right)}^{b \left({y}\right)} f \left({x, y}\right) \rd x = f \left({y, b \left({y}\right)}\right) \frac {\d b} {\d y} - f \left({y, a \left({y}\right)}\right) \frac {\d a} {\d y} + \int_{a \left({y}\right)}^{b \left({y}\right)} \frac {\partial} {\partial y} f \left({x, y}\right) \rd x$

where $a \left({y}\right)$, $b \left({y}\right)$ are continuously differentiable.

Setting $a \left({y}\right)$ and $b \left({y}\right)$ constant, so that $\exists a, b \in \R: a \left({y}\right) = a, b \left({y}\right) = b$:

$\dfrac {\d a} {\d y} = \dfrac {\d b} {\d y} = 0$

from which the result follows immediately.

$\blacksquare$


Proof 2

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

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

Then by linearity of the integral:

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

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 $\displaystyle \frac {\d f_y} {\d x} = \frac {\partial f} {\partial x}$.

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

That is:

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

Therefore:

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


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:

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

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

$\blacksquare$