# Integration by parts hint

1. Dec 7, 2005

### 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

Last edited: Dec 7, 2005
2. Dec 7, 2005

### George Jones

Staff Emeritus
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

3. Dec 7, 2005

### Ratzinger

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