Difference between revisions of "Integration by parts for delta integrals with no sigma in integrand"

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(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=...")
 
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==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.

Proof

References