# Evaluation of Riemann-Stieltjes Integral

I need to find the value of $\int_2^7 t^2 d\alpha(t)$ where $\alpha(t)$ is defined by the condition that $\alpha(t)=-t+k$ for $k\le t \lt k+1$

My textbook (Real Analysis and Foundations by S. Krantz) evaluates the Riemann-Stieltjes integral by determining the infimum of upper Riemann sum and the supremum lower Riemann sum. Therefore, I would like to evaluate the integral as the method in the textbook.

My hunch tells me that since $\alpha'(t)=-1$ for every interval $(k, k+1)$, the value of the integral is the same as $-\int_2^7 t^2 dt$ which is $-\frac{335}{3}$. However, I am having trouble dealing with $\alpha(t)$ since it is not monotone.

I have thought of dividing the integral into $\int_2^3 t^2 d\alpha(t)$, $\int_3^4 t^2 d\alpha(t)$, ... and evaluate them separately and add them up. But I have no clue why this is possible (if this is possible). To be exact, I am not sure how to deal with the discontinuities in $\alpha(t)$.

Thank you.

• I have $1$ doubt. If nothing is said about $k$ , why can't we take $k=2$ ? (Since you said for every interval $(k,k+1)$ ) Nov 24, 2016 at 15:48
• And if we can, then I think , it is the reason why your thought will work ! Nov 24, 2016 at 15:49
• @Qwerty How should I treat the endpoints of each intervals? I mean, how can I prove that the value of the integral doesn't matter whether $\alpha(t)$ is continuous at the endpoints or not. Nov 24, 2016 at 15:58
• Since $d\alpha$ is given to be valid between $t\in(2,7)$, I think it is safe to assume that $\alpha(t)$ is continuous at least in that interval.(otherwise $d\alpha$ and hence the entire integral wouldn't exist!) Nov 24, 2016 at 16:04
• $\alpha$ is not continuous. The jump at an integer $k$ will add $k^2$ to the integral. Nov 24, 2016 at 16:27

Jumps of $\alpha$ manifest as point masses in $d\alpha$.

Intuitively, $d\alpha$ is something that puts mass $\Delta \alpha = \alpha(x+\Delta x) - \alpha(x)$ on the interval $[x,x+\Delta x]$. If $\Delta x$ is small and $\alpha$ is differentiable at $x$, then $\Delta \alpha \approx \alpha'(x) \, \Delta x$ is the length of interval multiplied by the factor $\alpha'(x)$. This represents the situation where the mass is well-spread over the interval $[x,x+\Delta x]$ with density $\alpha'(t)$. On the other hand, if $\alpha$ has jump of size $m$, then $\Delta \alpha \approx m$ regardless of the width of the interval $[x, x+\Delta x]$. This means that the mass of $m$ is concentrated at a single point $x$.

As to the actual computation, we can utilize the observation that your $\alpha$ can be conveniently written as $\alpha(t) = \lfloor t \rfloor - t$ to write

$$\int_{2}^{7} t^2 \, d\alpha(t) = \int_{2}^{7} t^2 \, d\lfloor t \rfloor - \int_{2}^{7} t^2 \, dt$$

For the former, $t \mapsto \lfloor t \rfloor$ increases only at its jumps on $[2, 7]$, which are precisely $t = 3, 4, 5, 6, 7$. Since the jump sizes are identically 1,

$$\int_{2}^{7} t^2 \, d\lfloor t \rfloor = 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 135.$$

For the latter, the computation is straightforward since it is now a usual Riemann integral. So

$$\int_{2}^{7} t^2 \, dt = \frac{335}{3}.$$

Evaluate the integral on each interval $(k,k+1)$, and this gives what you have. Then add $k^2$ for each jump: $3^2+4^2+5^2+6^2+7^2$.

To explain why, look back at the definition of the Riemann-Stieltjes integral $\int_a^bf(t)dg(t)$. Around a jump, there will be some interval $(x_i,x_{i+1})$ and the contribution to the integral will be $f(x_i^*)(g(x_{i+1})-g(x_i))$, where $x_{i}^*$ is in $(x_i,x_{i+1})$. As the length of the interval shrinks to zero, the expression limits to the size of the jump times the value of $f$ at the location of the jump.