# Compute the Riemann-Stieltjes integral $\int_{-1}^1 \cos x\ \mathsf dg(x)$ without reduction.

Compute the Riemann-Stieltjes integral $$\int_{-1}^{1}\cos x \ \mathsf dg(x),$$ where $$g(x) = -\mathsf 1_{[-1,0]}(x)+3\cdot \mathsf 1_{(0,1]}(x)$$.

Solution: $$\int_{-1}^{1}\cos x\ \mathsf dg(x) = f(c)(g(c^{+})-g(c^{-})=\cos(0)(g(0^{+})-g(0^{-}))=1*(3-(-1))=4$$

Could someone tell me if I have solved this correctly?

• Yes, $4$ is the correct answer. Jan 23, 2020 at 5:16

Your answer is correct. Rigorously, if we take any partition $$-1=x_0, then the nature of $$g$$ dictates that in the expression for the upper or lower Riemann-Stieltjes sum , only that interval contributes which contains $$0$$ ; that term will look like $$\inf_{[x_j,x_{j+1}]}[\cos(x)][3-(-1)] = 4(\inf_{[x_j,x_{j+1}]}\cos(x))$$ (or $$\sup$$ for the upper RS sum). As $$\cos$$ is continuous, this term converges to $$4 \times \cos 0 = 4$$ for the upper and lower case.
In this case $$g$$ is differentiable except at one point where it has a jump discontinuity : for these kind of functions it is possible to create a general formula for the RS integral in terms of the derivative, and the amount of jump at the point. One can express the jump unrigorously as a "delta function" contribution in the derivative, and proceed to evaluate as if $$dg(x) = g'(x)dx + \delta$$ term. In this case, $$g' \equiv 0$$ when defined and the value of the jump is $$4$$ at the point $$0$$, so we get $$dg(x) = 0dx + 4\delta_0$$, and upon integration, only the delta remains : $$4\delta_0(\cos(x)) = 4 \cos 0 = 4$$.