The Formula for Integration by Parts of Riemann-Stieltjes Integrals

# The Formula for Integration by Parts of Riemann-Stieltjes Integrals

We have looked at some nice properties regarding Riemann-Stieltjes integrals on the following pages:

We will now look at another very important relationship between a two functions $f$ and $\alpha$ that are both Riemann-Stieltjes integrable with respect to one another on the interval $[a, b]$.

 Theorem 1: Let $f$ be a Riemann-Stieltjes integrable function with respect to $\alpha$ on the interval $[a, b]$. Then $\alpha$ be a Riemann-Stieltjes integrable function with respect to $f$ on the interval $[a, b]$ and $\displaystyle{\int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) = f(b)\alpha(b) - f(a)\alpha(a)}$.
• Proof: To show that $\displaystyle{\int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) = f(b)\alpha(b) - f(a)\alpha(a)}$ we will show that for all $\epsilon > 0$ then:
(1)
\begin{align} \quad \biggr \lvert \left ( \int_a^b f(x) \: d\alpha(x) + \int_a^b \alpha (x) \: d f(x) \right ) - [f(b)\alpha(b) - f(a)\alpha(a)] \biggr \rvert < \epsilon \end{align}
• Let $\epsilon > 0$ be arbitrary given. Now since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on the interval $[a, b]$ then for some $A \in \mathbb{R}$, $\int_a^b f(x) \: d \alpha (x) = A$, and for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, b]$ such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_1}$ ($P_{\epsilon_1} \subseteq P$) and for any choice of $t_k$ is each $k^{\mathrm{th}}$ subinterval we have that:
(2)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• Similarly, since $\alpha$ is Riemann-Stieltjes integrable with respect to $f$ on the interval $[a, b]$ then for some $B \in \mathbb{R}$, $\int_a^b \alpha (x) \: d f(x) = B$, and for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a partition $P_{\epsilon_2} \in \mathscr{P}[a, b]$ such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_2}$ ($P_{\epsilon_2} \subseteq P$) and for any choice of $u_k$ in each $k^{\mathrm{th}}$ subinterval we have that:
(3)
\begin{align} \quad \mid S(P, \alpha, f) - B \mid < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}
• Let $P_{\epsilon} = P_{\epsilon_1} \cup P_{\epsilon_2}$. Then for all partitions $P = \{a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ finer than $P_{\epsilon}$ ($P_{\epsilon} \subseteq P$), we have that $(*)$ and $(**)$ are satisfied, so:
(4)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid + \mid S(P, \alpha, f) - B \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \\ \quad \mid S(P, f, \alpha) + S(P, \alpha, f) - (A + B) \mid < \epsilon \end{align}
• Now choose $t_k = x_k$ and $u_k = x_{k-1}$ for each $k \in \{1, 2, ..., n \}$. Then we have that:
(5)
\begin{align} \quad \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta \alpha_k + \sum_{k=1}^{n} \alpha(u_k) \Delta f_k - (A + B) \biggr \rvert < \epsilon \\ \quad \biggr \lvert \sum_{k=1}^{n} \left [ f(x_k) \Delta \alpha_k + \alpha(x_{k-1}) \Delta f_k \right ] - (A + B) \biggr \rvert < \epsilon \\ \quad \biggr \lvert \sum_{k=1}^{n} \left [ f(x_k) [\alpha (x_k) - \alpha(x_{k-1})] + \alpha(x_{k-1})[f(x_k) - f(x_{k-1}) \right ] - (A + B) \biggr \rvert < \epsilon \\ \quad \biggr \lvert \sum_{k=1}^{n} \left [ f(x_k)\alpha(x_k) - f(x_k)\alpha(x_{k-1}) + \alpha(x_{k-1})f(x_k) - \alpha(x_{k-1})f(x_{k-1}) \right ] - (A + B) \biggr \rvert < \epsilon \\ \quad \biggr \lvert \sum_{k=1}^{n} \left [ f(x_k)\alpha(x_k) - f(x_{k-1})\alpha(x_{k-1}) \right ] - (A + B) \biggr \rvert < \epsilon \\ \end{align}
• The finite series $\sum_{k=1}^{n} \left [ f(x_k)\alpha(x_k) - f(x_{k-1})\alpha(x_{k-1}) \right ] = f(x_n)\alpha(x_n) - f(x_0)\alpha(x_0) = f(b)\alpha(b) - f(a)\alpha(a)$ as a telescoping series, so:
(6)
\begin{align} \quad \mid f(b)\alpha(b) - f(a)\alpha(a) - (A + B) \mid < \epsilon \\ \quad \biggr \lvert \left ( \int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) \right ) - [f(b)\alpha(b) - f(a)\alpha(a)] \biggr \rvert < \epsilon \end{align}
• But $\epsilon > 0$ is arbitrary, and so:
(7)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) = f(b)\alpha(b) - f(a)\alpha(a) \quad \blacksquare \end{align}