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Differentiating $\int_{0}^{\infty} e^{- t x} \sin x d x$ under the integral sign

Example

For $F (t) : = \int_{0}^{\infty} e^{- t x} \sin x d x (t > 0),$ differentiation under the integral sign is legal, and $F^{'} (t) = - \int_{0}^{\infty} x e^{- t x} \sin x d x .$

Facts & Assumptions

Given: The parameter integral $F (t)$ for $t > 0$ .

[L1]

Differentiation under the integral sign is valid under an integrable dominating bound for the parameter derivative (Differentiation under the integral sign).

Verification

technique · direct

1.1

Fix a compact interval $[a, b] \subset (0, \infty)$ . For $f (x, t) : = e^{- t x} \sin x,$ one has $\frac{\partial f}{\partial t} (x, t) = - x e^{- t x} \sin x,$ so $∣ \frac{\partial f}{\partial t} (x, t) ∣ \leq x e^{- a x} (t \in [a, b]) .$

givenconstructalgebra

2.1

The function $x \mapsto x e^{- a x}$ is integrable on $[0, \infty)$ , so [L1] applies on every compact parameter interval and yields $F^{'} (t) = - \int_{0}^{\infty} x e^{- t x} \sin x d x .$ This is exactly the advertised differentiation step.

step 1.1L1∎

Depends on

Differentiation under the integral sign

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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)

Differentiating ∫0∞e−txsin⁡x dx∫0∞​e−txsinxdx under the integral sign