TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Differentiation under the integral sign

Statement

Let $I \subseteq R$ be an open interval and let $f : X \times I \to C$ be such that:

for every $t \in I$ , the function $x \mapsto f (x, t)$ is integrable;
for almost every $x$ , the map $t \mapsto f (x, t)$ is differentiable on $I$ ;
for every $t \in I$ , the function $x \mapsto \frac{\partial f}{\partial t} (x, t)$ is measurable;
there are a measurable null set $N$ and a nonnegative measurable function $g$ with $\int g d μ < + \infty$ and $∣ \frac{\partial f}{\partial t} (x, t) ∣ \leq g (x)$ for every $t \in I$ and every $x \in X ∖ N$ .

Then $F (t) : = \int f (x, t) d μ (x)$ is differentiable on $I$ , and $F^{'} (t) = \int \frac{\partial f}{\partial t} (x, t) d μ (x) .$

Facts & Assumptions

Given: An open interval $I$ , a function $f$ satisfying the first three displayed hypotheses, and a measurable null set $N$ together with a nonnegative measurable majorant $g$ satisfying hypothesis 4.

[L1]

Dominated convergence applies to integrable complex-valued functions under a single $L^{1}$ majorant (Dominated convergence).

[L2]

The mean value theorem bounds difference quotients by a derivative bound on an interval (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · direct

1.1

Fix $t_{0} \in I$ and let $h_{n} \to 0$ with $h_{n} \neq 0$ and $t_{0} + h_{n} \in I$ . Define $q_{n} (x) : = \frac{f (x, t_{0} + h_{n}) - f (x, t_{0})}{h_{n}} .$ For every $x \in X ∖ N$ , differentiability in $t$ gives $q_{n} (x) \to \partial_{t} f (x, t_{0})$ .

givenconstruct

2.1

For each $n$ , hypothesis 1 makes $x \mapsto f (x, t_{0} + h_{n})$ and [step 1.1, L1, L2] $x \mapsto f (x, t_{0})$ integrable and therefore measurable, so $q_{n}$ is measurable. Fix $x \in X ∖ N$ and $n$ . If $q_{n} (x) = 0$ , then $∣ q_{n} (x) ∣ \leq g (x)$ is immediate. Otherwise put $α : = \overline{q_{n} (x)} / ∣ q_{n} (x) ∣,$ so $∣ α ∣ = 1$ and $∣ q_{n} (x) ∣ = Re ⁣ (α \frac{f (x, t_{0} + h_{n}) - f (x, t_{0})}{h_{n}}) .$ Apply [L2] to the real-valued function $τ \mapsto Re (α f (x, τ))$ on the segment joining $t_{0}$ to $t_{0} + h_{n}$ . For some interior point $ξ$ of that segment, $∣ q_{n} (x) ∣ = Re (α \partial_{t} f (x, ξ)) \leq ∣ \partial_{t} f (x, ξ) ∣ \leq g (x) .$ Hypothesis 3 makes $x \mapsto \partial_{t} f (x, t_{0})$ measurable. Therefore [L1] applies to $(q_{n})$ .

step 1.1L1L2

3.1

By [L1], $\lim_{n} \int q_{n} (x) d μ (x) = \int \partial_{t} f (x, t_{0}) d μ (x) .$ But $\int q_{n} (x) d μ (x) = \frac{F (t_{0} + h_{n}) - F (t_{0})}{h_{n}},$ so the difference quotients of $F$ converge to the displayed integral. Hence $F$ is differentiable at $t_{0}$ with the stated derivative.

step 2.1L1algebra∎

Depends on

Used by

Differentiating ∫₀^∞ e⁻ᵗˣsin x dx under the integral sign Example

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.27 (standard reference, not scraped)