Difference between revisions of "Delta integral from t to sigma(t)"
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(Created page with "==Theorem== The following formula holds: $$\int_t^{\sigma(t)} f(\tau) \Delta \tau = \mu(t)f(t)$$ ==Proof== ==References== Category:Theorem Category:Unproven") |
m (Tom moved page Integral from t to sigma(t) to Delta integral from t to sigma(t)) |
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$\int_t^{\sigma(t)} f(\tau) \Delta \tau = \mu(t)f(t)$$ | + | $$\int_t^{\sigma(t)} f(\tau) \Delta \tau = \mu(t)f(t),$$ |
| + | where $\int$ denotes the [[delta integral]], $\sigma$ denotes the [[forward jump]], and $\mu$ denotes the [[forward graininess]]. | ||
==Proof== | ==Proof== | ||
Latest revision as of 22:40, 22 August 2016
Theorem
The following formula holds: $$\int_t^{\sigma(t)} f(\tau) \Delta \tau = \mu(t)f(t),$$ where $\int$ denotes the delta integral, $\sigma$ denotes the forward jump, and $\mu$ denotes the forward graininess.
