Prove that for functions $u(x)$ and $v(x)$ which are $n+1$ times differentiable on $[a, b]$ the following equality holds:
$$
\int_a^b uv^{(n+1)}dx = \left.\sum_{k=0}^n (-1)^ku^{(k)}v^{(n-k)}\right\vert_a^b + (-1)^{n+1}\int_a^bu^{(n+1)}v\;dx
$$
I've been proving a simpler case for the "first-order" integration by parts formula for proper integrals, namely:
$$
\int_a^b u\;dv = \left.uv\right|_a^b + \int_a^bv\;du
$$
I did this by considering the following differential:
$$
d(uv) = d(u)v + vd(u)
$$
And then integrating both parts on $[a, b]$:
$$
\int_a^bd(uv) = \int_a^bu\;d(v)+\int_a^bv\;d(u)\\
\left.uv\right|_a^b = \int_a^bu\;d(v)+\int_a^bv\;d(u)\\ \implies
\int_a^bu\;d(v) = \left.uv\right|_a^b – \int_a^bv\;d(u)
$$
I then tried to use the same approach for the "higher-order" formula but failed. It's not difficult to show that:
$$
(uv)^{(n)} = \sum_{k=0}^n {n\choose k} u^{(n-k)}v^{(k)}
$$
Where $(f(x))^{(n)}$ denotes $n$-th derivative of $f(x)$.
I believe this is the basic idea for the proof, but unfortunately, I do not see how to proceed from here. I would appreciate any help with this question. Thank you!
Best Answer
It can be proven quite straightforwardly by induction. We know the base case, for $1$ iteration of integration by parts (IBP) as you have proven it. So assuming $\displaystyle \int uv^{(n+1)} = \sum_{k=0}^n (-1)^ku^{(k)}v^{(n-k)} + (-1)^{n+1}\int_a^bu^{(n+1)}v$, for some $n\in\mathbb{N_0}$, we would have
$$ \begin{align} \int uv^{(n+\color{red}{2})} &=uv^{(n+1)}-\int u'v^{(n+1)} \\ &=uv^{(n+1)}-\sum_{k=0}^n (-1)^k(u')^{(k)}v^{(n-k)} - (-1)^{n+1}\int_a^b(u')^{(n+1)}v \\ &=\left[uv^{(n+1)}-u^{(1)}v^{(n)}+\ldots+(-1)^{n+1}u^{(n+1)}v^{(0)}\right] + (-1)^{n+2}\int_a^bu^{(n+2)}v \\ &=\sum_{k=0}^{n+1} (-1)^ku^{(k)}v^{(n-k)} + (-1)^{n+2}\int_a^bu^{(n+2)}v\end{align}$$
Therefore, it is true for all $n\in\mathbb{N}_0$.