3.3. Integration by Parts

Theorem 3.4

If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), then \(\alpha \in \mathfrak{R}(f)\) on \([a, b]\), and

(3.5)\[\int_a^b f \dif\alpha + \int_a^b \alpha\dif f = f(b) \alpha(b) - f(a) \alpha(a)\]

Note

Take a second and appreciate the beauty of symmetry of the equation (3.5). This can be regarded as a reciprocal rule for Riemann-Stieltjes integrals. Indeed, it tells us the value of the integral when the integrand and the integrator are swapped.

Proof. Let \(\varepsilon > 0\) be arbitrary. Because \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), by definition, there exists \(P_\varepsilon\) such that for any refinement \(P \supseteq P_\varepsilon\) and any set of representatives \(T\) of \(P\), the Riemann-Stieltjes sum \(S(P, T, f, \alpha)\) satisfies that

\[\abs{S(P, T, f, \alpha) - \int_a^b f \dif \alpha} < \varepsilon\]

Consider an arbitrary refinement \(P^\prime \supseteq P_\varepsilon\). And let \(T^\prime\) be a list of representatives of \(P^\prime\). We want to show that \(S(P^\prime, T^\prime, \alpha, f)\) is near the desired value. Write \(P^\prime = \{x_0, \ldots, x_n\}\). The Riemann-Stieltjes sum \(S(P^\prime, T^\prime, \alpha, f)\) can be then written as

(3.6)\[S(P^\prime, T^\prime, \alpha, f) = \sum_{k=1}^n \alpha(t_k) [f(x_k) - f(x_{k-1}] = \sum_{k=1}^n \alpha(t_k) f(x_k) - \sum_{k=1}^n \alpha(t_k) f(x_{k-1})\]

Meanwhile, the difference \(A = f(b)\alpha(b) - f(a)\alpha(a)\) on the right-hand side of (3.5) can be written as

(3.7)\[\begin{split}A & = & & f(b)\alpha(b) - f(a)\alpha(a) \nonumber\\& = & & f(x_n)\alpha(x_n) - f(x_{n-1})\alpha(x_{n-1}) \nonumber\\& & & + f(x_{n-1})\alpha(x_{n-1}) - \cdots - f(x_{1})\alpha(x_{1}) \nonumber\\& & & + f(x_{1})\alpha(x_{1})- f(x_0)\alpha(x_0) \nonumber\\& = & & \sum_{k=1}^n f(x_k)\alpha(x_k) - \sum_{k=1}^n f(x_{k-1})\alpha(x_{k-1})\end{split}\]

Subtracting (3.6) from (3.7), we obtain

(3.8)\[A - S(P^\prime, T^\prime, \alpha, f) = \sum_{k=1}^n f(x_k) [\alpha(x_k) - \alpha(t_k) ] + \sum_{k=1}^n f(x_{k-1}) [\alpha(t_k) - \alpha(x_{k-1}) ]\]

Taking a close look at the right-hand side of (3.8), one may realize that it is also a Riemann-Stieltjes sum. To see this, let \(P^{\prime\prime} = P^\prime \cup T^\prime\), and let \(T^{\prime\prime}\) be the list of representatives constructed as follows. Choose \(x_k\) in \([t_k, x_k]\)(if \(t_k < x_k\)) and chose \(x_{k-1}\) in \([x_{k-1}, t_k]\)(if \(x_{k-1} < t_k\)).

Note

There are chances that \(t_k = x_k\) or \(t_k = x_{k-1}\). In that case, the term \(f(x_k)[\alpha(x_k)-\alpha(t_k)]\) or \(f(x_{k-1})[\alpha(t_k)-\alpha(x_{k-1})]\) would be zero.

Consider the diagram shown in Figure Fig. 3.1.

Fig. 3.1 The blue part is associated with \(f(x_k)[\alpha(x_k)-\alpha(t_k)]\) while the orange part is associated with \(f(x_{k-1})[\alpha(t_k)-\alpha(x_{k-1})]\). We see that summing them up yields \(S(P^{\prime\prime}, T^{\prime\prime}, f, \alpha)\).

The right-hand side of (3.8) is just \(S(P^{\prime\prime}, T^{\prime\prime}, f, \alpha)\). Since \(P^{\prime\prime} \supseteq P_\varepsilon\), we have

\[\abs{ S(P^{\prime\prime}, T^{\prime\prime}, f, \alpha) - \int_a^b f \dif \alpha } < \varepsilon\implies\abs{ A - S(P^\prime, T^\prime, \alpha, f) - \int_a^b f \dif \alpha } < \varepsilon\]

This shows that \(\alpha \in \mathfrak{R}(f)\) on \([a, b]\), and (3.5) is proved.