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Differentiation under an improper multiple integral under an integrable derivative bound

Statement

Let $D \subseteq R^{n}$ be open and $I \subseteq R$ be an open interval. Suppose $f$ and $\partial_{t} f$ are continuous on $D \times I$ , one slice $f_{t_{*}}$ is absolutely improperly integrable, and for every compact interval $C \subseteq I$ there is a nonnegative improperly integrable $g_{C}$ with $∣ \partial_{t} f (x, t) ∣ \leq g_{C} (x)$ for $x \in D$ and $t \in C$ . Then every slice is absolutely improperly integrable, the function $F (t) = \int_{D} f (x, t) d x$ is continuously differentiable, and

$F^{'} (t) = \int_{D} \partial_{t} f (x, t) d x .$

The parameter derivative may be passed through the improper multiple integral under an integrable uniform derivative bound.

Facts & Assumptions

Given: The domain, interval, integrand, derivative, base slice, and dominators of the Statement; fix $t_{0} \in I$ .

[L1]

The mean value theorem gives an interior $c$ with $h (b) - h (a) = h^{'} (c) (b - a)$ for a continuous function differentiable inside 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 $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L2]

If locally integrable slices $u_{t}$ satisfy $∣ u_{t} ∣ \leq g$ on a compact parameter set, where $g \geq 0$ and $\int_{D} g < + \infty$ , then their improper-integral tails outside one compact Jordan core are uniformly small (An integrable dominator gives uniform tail control on every compact parameter set).

[L3]

If $u : D \times I \to R$ is continuous and locally dominated near every parameter by a nonnegative improperly integrable function, then $t \mapsto \int_{D} u (x, t) d x$ is continuous in the relative topology (Locally dominated parameter-dependent improper multiple integrals are continuous).

[L4]

If $∣ u ∣ \leq g$ and $g$ has finite nonnegative improper integral, then $u$ is absolutely improperly integrable (Comparison and absolute comparison tests for improper multiple integrals).

[L5]

A continuous map from a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L6]

Proper multidimensional integrals are linear and satisfy $∣ \int u ∣ \leq \int ∣ u ∣$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L7]

If $u$ is absolutely improperly integrable, its proper integrals along every compact Jordan exhaustion converge to $\int_{D} u$ (Absolute convergence makes signed improper multiple integrals independent of exhaustion).

Proof

technique · direct

1.1

On the compact parameter interval between $t_{*}$ and any $t \in I$ , [L1] gives $∣ f (x, t) - f (x, t_{*}) ∣ \leq ∣ t - t_{*} ∣ g_{C} (x)$ pointwise. Fact [L4] makes the difference absolutely improperly integrable. Proper linearity in [L6] along one exhaustion and convergence in [L7] show that a sum of two absolutely improperly integrable functions is again absolutely improperly integrable, so $f_{t} = f_{t_{*}} + (f_{t} - f_{t_{*}})$ is absolutely improperly integrable.

L1L4L6L7algebra

1.2

For nonzero $h$ with $t_{0} + h \in I$ , [L1] bounds the difference quotient $q_{h} (x) : = (f (x, t_{0} + h) - f (x, t_{0})) / h$ by one integrable dominator. On a compact Jordan core, [L5] and [L1] make $q_{h} \to \partial_{t} f (\cdot, t_{0})$ uniformly as $h \to 0$ .

L1L2L5

2.1

Use [L2] to make the tails of both $q_{h}$ and $\partial_{t} f (\cdot, t_{0})$ uniformly small, then use the uniform convergence from step 1.2 on the core and [L6]. It follows that $\int_{D} q_{h} \to \int_{D} \partial_{t} f (x, t_{0}) d x$ . On every compact member of one exhaustion, proper linearity in [L6] gives $\int q_{h} = (\int f_{t_{0} + h} - \int f_{t_{0}}) / h$ ; applying [L7] to the three absolutely integrable functions passes this identity to $D$ . Hence the left side is the difference quotient of $F$ , proving the asserted derivative formula.

step 1.2L2L6L7

3.1

Apply [L4] to each derivative slice and [L3] to the continuous integrand $\partial_{t} f$ , using the same local dominators, to see that the derivative integral is continuous in $t$ . Thus $F$ is $C^{1}$ on $I$ .

step 2.1L3L4∎

Depends on

Used by

Differentiation under an improper integral can fail without uniform domination Counterexample
The scaled Gaussian integral and its parameter derivative Example

Dependency tree · two levels

40 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

W. F. Trench, Functions Defined by Improper Integrals, Theorem 11 (standard reference, not scraped)