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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $11.5$