Difference between revisions of "Integration by parts for delta integrals with no sigma in integrand"
From timescalewiki
(Created page with "==Theorem== The following formula holds: $$\int_a^b f(t) g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t) g(\sigma(t)) \Delta t.$$ ==Proof== ==References=...") |
|||
Line 1: | Line 1: | ||
==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
− | $$\int_a^b f(t) g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t) g(\sigma(t)) \Delta t | + | $$\int_a^b f(t) g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t) g(\sigma(t)) \Delta t,$$ |
+ | where $\int$ denotes the [[delta integral]]. | ||
==Proof== | ==Proof== |
Revision as of 23:26, 22 August 2016
Theorem
The following formula holds: $$\int_a^b f(t) g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t) g(\sigma(t)) \Delta t,$$ where $\int$ denotes the delta integral.