Leibniz's Integral Rule

Theorem
Let $\map f {x, t}$, $\map a t$, $\map b t$ be continuously differentiable real functions on some region $R$ of the $\tuple {x, t}$ plane.

Then for all $\tuple {x, t} \in R$:


 * $\ds \frac \d {\d t} \int_{\map a t}^{\map b t} \map f {x, t} \rd x = \map f {\map b t, t} \frac {\d b} {\d t} - \map f {\map a t, t} \frac {\d a} {\d t} + \int_{\map a t}^{\map b t} \frac \partial {\partial t} \map f {x, t} \rd x$

Proof
By the Mean Value Theorem for Integrals, there exist $\xi_a \in \closedint {\map a t} {\map a {t + h} }$ and $\xi_b \in \closedint {\map b t} {\map b {t + h} }$ such that:

Now, consider the last term:
 * $\ds \lim_{h \mathop \to 0} \int_{\map a t}^{\map b t} \frac {\map f {x, t + h} - \map f {x, t} } h \rd x$

By Limit of Function by Convergent Sequences, it suffices to find the following for an arbitrary sequence $\sequence {h_n}_{n \mathop \in \N}$ such that $\ds \lim_{n \mathop \to \infty} h_n = 0$:
 * $\ds \lim_{n \mathop \to 0} \int_{\map a t}^{\map b t} \frac {\map f {x, t + h_n} - \map f {x, t} } {h_n} \rd x$

Fix such as sequence, as well as a value for $t$.

For each $n \in \N$, define a function $\map {f_n} x$ as:
 * $\map {f_n} x = \frac {\map f {x, t + h_n} - \map f {x, t} } {h_n}$

when $\tuple {x, t}$ and $\tuple {x, t + h_n}$ are in $R$.

By the Extreme Value Theorem:
 * $M = \sup_{\tuple x \in R_t} \size {\frac \partial {\partial t} \map f {x, t} }$

is well-defined, where $R_t$ is the subset of $R$ where $t$ matches our chosen value.

Then, from the Mean Value Theorem, it follows that:
 * $\size {\map {f_n} x} \le M$

for all values of $n$ and $x$.

By Limit of Function by Convergent Sequences, as $n \to \infty$:
 * $\ds \map {f_n} x \to \lim_{h \mathop \to 0} \frac {\map f {x, t + h} - \map f {x, t} } h = \frac \partial {\partial t} \map f {x, t}$

by definition of Partial Derivative.

Therefore, by Lebesgue's Dominated Convergence Theorem:
 * $\ds \lim_{n \mathop \to 0} \int_{\map a t}^{\map b t} \frac {\map f {x, t + h_n} - \map f {x, t} } {h_n} \rd x = \int_{\map a t}^{\map b t} \frac \partial {\partial t} \map f {x, t} \rd x$

Combining with the result from before:
 * $\ds \frac \d {\d t} \int_{\map a t}^{\map b t} \map f {x, t} \rd x = \frac {\d b} {\d t} \map f {\map b t, t} - \frac {\d a} {\d t} \map f {\map a t, t} + \int_{\map a t}^{\map b t} \frac \partial {\partial t} \map f {x, t} \rd x$

Also see

 * Definite Integral of Partial Derivative, where $\map a t$ and $\map b t$ are constant:


 * $\ds \frac {\rd} {\rd t} \int_a^b \map f {x, t} \rd x = \int_a^b \frac {\partial} {\partial t} \map f {x, t} \rd x$