Difference between revisions of "Integration by parts for delta integrals with no sigma in integrand"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
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==Proof== | ==Proof== | ||
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| + | ==See also== | ||
| + | [[Integration by parts for delta integrals with sigma in integrand]]<br /> | ||
==References== | ==References== | ||
Latest revision as of 14:48, 13 March 2018
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
See also
Integration by parts for delta integrals with sigma in integrand
