Improper Integral of Partial Derivative

From ProofWiki
Jump to navigation Jump to search


Let $I$ and $U$ be open intervals of $\R$, and $f: U \times I \to \R$ be a function such that:

  1. for all $u \in U$, $f(\cdot,u) : I \to \R$ is integrable;
  2. for all $t \in I$, $f(t,\cdot) : U \to \R$ is $\CC^1$;
  3. there is an integrable function $\phi : I \to \R$ such that for all $(t,u) \ in I \times U$,
$\ds \size {\partial_u f(t,u)} \leq \phi(t)$

Then the function $F : U\to \R$ defined by

$\ds F(u) = \int_I f(t,u) \rd t$

is $\CC^1$ and

$\ds F'(u) = \int_I \partial_u f (t,u) \rd t$

In other words

$\ds \frac \d {\d u} \int_I f(t,u) \rd t = \int_I \partial_u f(t,u) \rd u$

for all $u \in U$.

This is a straightforward corollary of the following measure theoretic theorem.

Let $X \subset \R$ be an open interval, and $\Omega$ a measure space.

If $f : X \times \Omega \to \R$ is a function satisfying:

  1. for all $x \in X$, the function $f(x,\cdot) : \Omega \to \R$ is Lebesque integrable;
  2. for almost all $\omega \in \Omega$, the partial derivative $\partial_x f$ along $X$ of $x$ is defined
  3. there is an integrable function $\phi: X \to \R$ such that for all $x \in X$, and for almost all $\omega \in \Omega$,
$\ds \size {\partial_x f(x,\omega)} \leq \phi(x)$

Then the function $F : X \to \R$ defined as

$\ds F(x) = \int_\Omega f(x,\omega) \rd \omega$

has a derivative at all point $x\in I$ given by the identity

$\ds F'(x) = \int_\Omega \partial_x f(x,\omega) \rd \omega $

More concisely:

$\ds \frac \d {\d x} \int_\Omega f(x,\omega) \rd \omega = \int_\Omega \partial_x f(x,\omega) \rd \omega$

for all $x\in X$.