Evaluating Riemann-Stieltjes Integrals

# Evaluating Riemann-Stieltjes Integrals

We will now look at evaluating some Riemann-Stieltjes integrals. Before we do, be sure to recall the results summarized below. Let $f$ and $g$ be Riemann-Stieltjes integrable function with respect to $\alpha$ and $\beta$ on the interval $[a, b]$ and let $c \in \mathbb{R}$.

• Additivity of the Integrand: $\displaystyle{\int_a^b [f(x) + g(x)] \: d \alpha (x) = \int_a^b f(x) \: d \alpha (x) + \int_a^b g(x) \: d \alpha (x)}$.
• Homogeneity of the Integrand: $\displaystyle{\int_a^b cf(x) \: d \alpha (x) = c \int_a^b f(x) \: d \alpha (x)}$.
• Additivity of the Integrator: $\displaystyle{\int_a^b f(x) \: d [\alpha (x) + \beta(x)] = \int_a^b f(x) \: d \alpha (x) + \int_a^b f(x) \: d \beta (x)}$.
• Homogeneity of the Integrator: $\displaystyle{\int_a^b f(x) \: d[c \alpha(x)] = c \int_a^b f(x) \: d \alpha (x)}$.
• Integration by Parts: $\displaystyle{\int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) = f(b)\alpha(b) - f(a)\alpha(a)}$.
• Reduction to a Riemann-Integral: If $f$ is bounded on $[a, b]$, $\alpha'$ exists and is continuous on $[a, b]$ then $\displaystyle{\int_a^b f(x) \: d \alpha(x) = \int_a^b f(x) \alpha'(x) \: dx}$

Let's now look at some examples.

## Example 1

Evaluate the Riemann-Stieltjes integral $\int_0^1 x \: d x^2$.

We see that the integrand $f(x) = x$ is bounded on $[0, 1]$, the derivative of the integrator $\alpha (x) = x^2$ exists and is $\alpha'(x) = 2x$ and is continuous on $[0, 1]$, so we reduce the Riemann-Stieltjes integral above to get:

(1)
\begin{align} \quad \int_0^1 x \: dx^2 = \int_0^1 2x^2 \: dx = \frac{2}{3} \left [ x^3 \right ]_{0}^1 = \frac{2}{3} \end{align}

## Example 2

Evaluate the Riemann-Stieltjes integral $\int_0^{\pi} x \: d \cos x$.

Using integration by parts gives us that:

(2)
\begin{align} \quad \int_0^{\pi} x \: d \cos x + \int_0^{\pi} \cos x \: dx &= \pi \cos (\pi) - 0(\cos (0)) \\ \quad \int_0^{\pi} x \: d \cos x + \left [ \sin x \right ]_{x=0}^{x=\pi} &= -\pi \\ \quad \int_0^{\pi} x \: d \cos x &= - \pi \end{align}

## Example 3

Evaluate the Riemann-Stieltjes integral $\int_0^{\pi} (x + 1) \: d (\sin x + \cos x)$.

Using the additivity of the integrand we have that:

(3)
\begin{align} \quad \int_0^{\pi} (x + 1) d \: (\sin x + \cos x) = \int_0^{\pi} x \: d (\sin x + \cos x) + \int_0^{\pi} 1 \: d (\sin x + \cos x) \end{align}

Using the additivity of the integrator and we have that:

(4)
\begin{align} \quad \int_0^{\pi} (x + 1) d \: (\sin x + \cos x) = \int_0^{\pi} x \: d \sin x + \int_0^{\pi} x \: d \cos x + \int_0^{\pi} 1 \: d \sin x + \int_0^{\pi} 1 \: d cos x \end{align}

The Riemann-Stieltjes integral $\int_0^{\pi} x \: d \cos x = -\pi$ from example 1. The other three integrals can be evaluated by using Integration by parts:

(5)
\begin{align} \quad \int_0^{\pi} x \: d \sin x + \int_0^{\pi} \sin x \: dx &= \pi \sin (\pi) - 0 \sin (0) \\ \quad \int_0^{\pi} x \: d \sin x - [\cos x]_{0}^{\pi} &= 0 \\ \quad \int_0^{\pi} x \: d \sin x - (-1 -1) &= 0 \\ \quad \int_0^{\pi} x \: d \sin x &= - 2 \end{align}
(6)
\begin{align} \quad \int_0^{\pi} 1 \: d \sin x + \int_0^{\pi} \sin x \: d 1 &= 1 \sin (\pi) - 1 \sin (0) \\ \quad \int_0^{\pi} 1 \: d \sin x &= 0 \end{align}
(7)
\begin{align} \quad \int_0^{\pi} 1 \: d \cos x + \int_0^{\pi} \cos x \: d 1 &= 1 \cos (\pi) - 1 \cos (0) \\ \quad \int_0^{\pi} 1 \: d \cos x &= -1 - 1 \\ \quad \int_0^{\pi} 1 \: d \cos x &= -2 \end{align}

Therefore:

(8)
\begin{align} \quad \int_0^{\pi} (x + 1) \: d (\sin x + \cos x) = -2 - \pi+ 0 - 2 = -(4 + \pi) \end{align}