integration by parts

[Ratzinger]

this no homework, but nevertheless can someone hint me how this integration by parts works?

\int {d^4 } x\frac{{\partial L}}{{\partial \left( {\partial _\mu \phi } \right)}}\partial _\mu (\delta \phi ) = {\rm{ }} - \int {d^4 } x\partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}} \right)\delta \phi {\rm{ }} + {\rm{ }}\int {d^4 } x{\rm{ }}\partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi } \right)

where
L(\phi ,\partial _\mu \phi )


I don't understand where the second term on the RHS comes from. I thought the second term should be
\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi \left| {^b _a } \right. = 0


thanks

[George Jones]

This is just the product rule, i.e.,


\frac{{\partial L}}{{\partial \left( {\partial _\mu \phi } \right)}}\partial _\mu (\delta \phi ) + \partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}} \right)\delta \phi {\rm{ }} = \partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi } \right)


Regards,
George

[Ratzinger]

I knew that it was going to be simple. Thanks George!