Alphabeta Math
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Differentiating ∫0∞e−txsin⁡x dx under the integral sign

Example

For F(t):=∫0∞e−txsin⁡x dx(t>0), differentiation under the integral sign is legal, and F′(t)=−∫0∞xe−txsin⁡x dx.

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
1.1givenconstructalgebra

Fix a compact interval [a,b]⊂(0,∞). For f(x,t):=e−txsin⁡x, one has ∂f∂t(x,t)=−xe−txsin⁡x, so ∣∂f∂t(x,t)∣≤xe−ax(t∈[a,b]).

2.1step 1.1L1∎

The function x↦xe−ax is integrable on [0,∞), so [L1] applies on every compact parameter interval and yields F′(t)=−∫0∞xe−txsin⁡x dx. This is exactly the advertised differentiation step.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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