0

Consider the integral $$ I(a,b,c,d) := \int_c^d dy \int_a^b dx\; f'(y-x) g(x+y) $$ where I would like to apply integration by parts on the $x$ variable. My thinking is that I can straightforwardly apply integration by parts by defining $v'(x):=f'(y-x)$ and $u(x):= g(x+y)$ which implies that $v(x) = -f(y-x)$ and $u'(x):= g'(x+y)$, which would naively imply $$ I(a,b,c,d) = \int_c^d dy \bigg[ -f(y-x) g(x+y) \bigg]_{x = a}^{x = b} - \int_c^d dy \int_a^b dx\; \big[ -f(y-x) \big] g'(x+y) \ . \ \ \ \ \ (**) $$

However this link seems to imply that extra care needed to be taken with the boundary terms in the above. Is (**) correct here?

QuantumEyedea
  • 2,805

1 Answers1

2

The other link is dealing with a more general situation than you have here.

$$\int_c^d dy \int_a^b dx\; f'(y-x) g(x+y)$$

Literally means the integral $\int_c^d dy\;h(y)$, where the function $h$ is defined by $$h(y) = \int_a^bdx\; f'(y-x) g(x+y)$$ And in performing that integration, $y$ is a constant, so $f'(y-x)$ and $g(x + y)$ are, for its sake, just functions of $x$. You have applied integration by parts perfectly to show that $$h(y) = \bigg[ -f(y-x) g(x+y) \bigg]_{x = a}^{x = b} - \int_a^b dx\; \big[ -f(y-x) \big] g'(x+y)$$ for every $y$. And therefore the rest of your expression follows naturally.

Paul Sinclair
  • 43,643