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In a physics proof I've been following, the author claims that by using integration by parts on the LHS of the following equation n times, we get the RHS.

$$ \int_a^b f^{(n)}(x)g(x) dx = (-1)^{(n)}\int_a^b f(x)g^{(n)}(x) dx$$

I've been staring at this for hours and still don't understand why this is true. How would I go about proving this?

connorp
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  • There must be some boundary conditions that make $\left.f^{(k)}(x)g^{(n-k)}(x)\right|_a^b=0$. –  Nov 17 '15 at 02:21
  • This statement is only true if a right combination of the function and its derivatives are zero at the boundaries . – Paul Nov 17 '15 at 02:21

1 Answers1

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This is true with the correct boundary conditions. The easiest proof is inductive.

Base case:

$$\int_a^b f'(x) g(x) \mathrm{d}x = fg|_a^b - \int_a^b f(x) g'(x)\mathrm{d}x$$ in which case your result is valid if $f(b)g(b) =f(a)g(a) $

Inductive step:

\begin{align} \int_a^b f^{(n+1)} g ~\mathrm{d}x&= f^{(n)} g|_a^b - \int_a^b f^{(n)} g' ~\mathrm{d}x \\ &\overset{(a)}= 0 - (-1)^n \int_a^b f (g')^{(n)}~\mathrm{d}x \\ &= (-1)^{n+1} \int_a^b f g^{(n+1)}~\mathrm{d}x \end{align}

where equality $(a)$ is due to the induction hypothesis.

The boundary conditions required are $f^{(k)}(b)g^{(n-k)} (b) = f^{(k)}(a)g^{(n-k)} (a)$ for all $k \in [0:n-1]$.

stochasticboy321
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  • Does the interval of integration must be finite or is this also valid when $a$ and/or $b$ goes to infinity? – Pinteco May 03 '24 at 08:23
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    Things also work with infinite intervals, in the appropriate sense. Recall that such 'improper integrals' are limits of integrals over finite intervals. For finite $u,v$, we can write $$ \int_u^v f^{(n)} g = \left.\sum_{k = 1}^{n-1} (-1)^{k+1} f^{(n - k)} g^{(k-1)}\right|_u^v + (-1)^{n}\int_u^v fg^{(n)}$$ (show this; in this case its better to explicitly develop a recursion rather than to induct, but in any case the idea is the same). Then take the appropriate limits, and things pop out. The boundary condition would require the sum to be $0$ once you take the limit. – stochasticboy321 May 03 '24 at 13:33