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I am considering an integral given in Synge (p. 27). This integral is,

$$ J = \int (y_i dx^i - \lambda(u) \omega du) $$

where $\lambda$ is a Lagrange multiplier, and $u$ is some parameter.

In the text it then states,

Applying a variation and integrating by parts, we get

$$ \delta J = [y_i \delta x^i] + \int (\delta y_i dx^i - \delta x^i dy_i - \omega \delta \lambda du - \lambda \frac{\partial \omega}{\partial x^i} \delta x^i du - \lambda \frac{\partial \omega}{\partial y_i} \delta y_i du)$$

Can anyone clear up why this is? I understand integration by parts (or thought I did!) but cannot seem to reproduce this result.

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They are applying integration by parts to the term $\int y_i\, \delta \mathrm dx^i$. It goes like this: $$ y_i\, \delta \mathrm dx^i = y_i\, \mathrm d\delta x^i = \mathrm d\left( y_i \delta x^i \right) - \delta x^i \mathrm dy_i \;. $$

The term where you integrate over a total differential gives the boundary contribution, $$ \int \mathrm d\left( y_i \delta x^i \right) = \left[ y_i \delta x^i \right] \;, $$ compare to the fundamental theorem of calculus $$ \int_a^b \frac{\mathrm df}{\mathrm dx}\, \mathrm dx = \int_{f(a)}^{f(b)} \mathrm df = \left. f \right|_a^b \;. $$

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    Could you be more explicit in the first line? I get you're integrating by parts, but where does the differential in the first term come from? – 2017-01-19
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    The differential is already in the expression you wrote in your question. If you vary the first term in the integrand, you get $\delta(y_i\, \mathrm dx^i) = \delta y_i\, \mathrm dx^i + y_i\, \delta \mathrm dx^i$ and I only considered the second of those terms because that's where the partial integration happens. – 2017-01-19
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    In the next step, $\mathrm d$ and $\delta$ commute by definition. In the last step of that line, note that "partial integration" is exactly the same thing as "product rule backwards". – 2017-01-19