Differentiability under Integral Sign

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\left({a \,.\,.\, b}\right)$ be a non-empty open interval.

Let $f: \left({a \,.\,.\, b}\right) \times X \to \R$ be a mapping satisfying:

$(1): \quad$ For all $t \in \left({a \,.\,.\, b}\right)$, the mapping $x \mapsto f \left({t, x}\right)$ is $\mu$-integrable
$(2): \quad$ For all $x \in X$, the mapping $t \mapsto f \left({t, x}\right)$ is differentiable
$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that:
$\displaystyle \forall \left({t, x}\right) \in \left({a \,.\,.\, b}\right) \times X: \left\vert{ \frac {\partial} {\partial t} f \left({t, x}\right)}\right\vert \le g \left({x}\right)$


Then the mapping $h: \left({a \,.\,.\, b}\right) \to \R$ defined by:

$\displaystyle h \left({t}\right) := \int f \left({t, x}\right) \, \mathrm d \mu \left({x}\right)$

is differentiable, and its derivative is:

$\displaystyle D_t \, h \left({t}\right) = \int \frac {\partial} {\partial t} f \left({t, x}\right) \, \mathrm d \mu \left({x}\right)$


Proof


Sources