calculation of Riemann–Stieltjes integral

• If $f$ is defined on  $[a,b]$  and $g$ is a constant function, then

  $${\int }_a^{b}f𝑑g=\mathrm{\hspace{0.33em}0}.$$

• Let $f$ be continuous on  $[a,b]$,    and  $g$ the step function defined as

  Then

  $${\int }_a^{b}f𝑑g=f(c)\cdot \alpha .$$

• Let $f$ be continuous on  $[a,b]$,    and the function $g$ be otherwise continuous but have in  $x=c$  a step of magnitude $\alpha$.  Then $g$ is sum of a continuous function $g^{*}$ and a step function

  and one has

  $${\int }_a^{b}f𝑑g={\int }_a^{b}fd(g^{*}+h)={\int }_a^{b}f𝑑g^{*}+{\int }_a^{b}f𝑑h={\int }_a^{b}f𝑑g^{*}+f(c)\cdot \alpha .$$

• Suppose that $g$ can be expressed in the form  $g=g^{*}+h$  where $g^{*}$ is continuous and $h$ a step function having an at most denumerable amount of steps ${\alpha }_i$ in respectively the same points $c_i$ on the interval  $[a,b]$  as the function $g$.  If $f$ is Riemann–Stieltjes integrable on  $[a,b]$,  then

  ${\displaystyle {\int }_a^b}f𝑑g={\displaystyle {\int }_a^b}f𝑑g^{*}+{\displaystyle \sum _i}f(c_i)\cdot {\alpha }_i.$ (1)

• Suppose that  $g=g^{*}+h$ (as above) has a finite amount of steps ${\alpha }_i$ in the points $c_i$ of the interval  $[a,b]$  but $f$ does not have same-sided discontinuities as $g$ in any of those points.  Then $f$ is Riemann–Stieltjes integrable on the interval and the equation (1) is true.

Example.  Find the value of the Riemann–Stieltjes integral

$$I:={\int }_{-3}^6(x-\lfloor x\rfloor )𝑑g(x)$$

where the integrand $f$ is the mantissa function and the integrator $g$ defined by

Now, $f$ is from the left discontinuous at every integer, but $g$ is bounded and only discontinuous from the right at $-2$ and 3.  By the above last item, $f$ is Riemann–Stieltjes integrable with respect to $g$ on  $[-3,\mathrm{\hspace{0.17em}6}]$.  We can set

$$g=g^{*}+h$$

where $g^{*}$ is continuous and the step function $h$ has the step of 2 at $-2$ and the step of 4 at 3.  Using (1) we get

$I$ $\mathrm{\hspace{0.33em}}={\displaystyle {\int }_{-3}^6}f𝑑g^{*}+f(-2)\cdot 2+f(3)\cdot 4={\displaystyle \sum _{i=-3}^5}{\displaystyle {\int }_i^{i+1}}f(x)g^{\prime }(x)𝑑x+0\cdot 2+0\cdot 4$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_{-3}^{-2}}(x+3)(-2x)𝑑x+{\displaystyle {\int }_{-2}^{-1}}(x+2)\cdot 1𝑑x+{\displaystyle {\int }_{-1}^0}(x+1)\cdot 1𝑑x+{\displaystyle {\int }_0^1}x\cdot 1𝑑x+{\displaystyle {\int }_1^2}(x-1)\cdot 1𝑑x$

$\mathrm{\hspace{1em}\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}+{\displaystyle {\int }_2^3}(x-2)\cdot 1𝑑x+{\displaystyle {\int }_3^4}(x-3)\cdot 2𝑑x+{\displaystyle {\int }_4^5}(x-4)\cdot 2𝑑x+{\displaystyle {\int }_5^6}(x-5)\cdot 2𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{47}{6}}.$

Title	calculation of Riemann–Stieltjes integral
Canonical name	CalculationOfRiemannStieltjesIntegral
Date of creation	2013-03-22 18:55:09
Last modified on	2013-03-22 18:55:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Topic
Classification	msc 26A42