What is the definite integral of $f(x)=x^2+1$ with respect to the differential of $\lfloor x\rfloor$ i.e ($d\lfloor x\rfloor$) from $0$ to $2$?
I tried to multiply and divide dx by then $d\lfloor x\rfloor/dx = 0$.
How do I approach it?
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do you mean $\int_0^2 (x^2+1)d\lfloor x\rfloor$? – Wouter Jul 12 '20 at 17:51
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@Wouter Well yes, we in India use that notation. – Harrison Wells Jul 12 '20 at 17:52
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$d\lfloor x\rfloor$ is not everywhere zero, in particular it is not zero at $x=1$. Are you familiar with the dirac delta? – Wouter Jul 12 '20 at 17:56
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@Wouter The dirac delta function? – Harrison Wells Jul 12 '20 at 17:58
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@Wouter Not that much but yes, please explain. – Harrison Wells Jul 12 '20 at 18:01
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2I think your question is about Riemann–Stieltjes integral ?! – Khosrotash Jul 12 '20 at 18:03
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Integrate by parts ( Why not ? ): \begin{align} &\bbox[#ffd,5px]{% \int_{0}^{2}\pars{x^{2} + 1}\dd\lfloor x\rfloor} \\[3mm] = &\ \overbrace{\left.\left\lfloor x\right\rfloor\pars{x^{2} + 1} \,\right\vert_{\ 0}^{\ 2}}^{\ds{=\ 10}}\ -\ \int_{0}^{2}\lfloor x\rfloor\pars{2x}\dd x \\[3mm] = &\ 10\ -\ \underbrace{2\int_{0}^{1}\lfloor x\rfloor\,\dd x}_{\ds{=\ 0}}\ -\ \underbrace{2\int_{1}^{2}\lfloor x\rfloor x\,\dd x}_{\ds{=\ 3}} = \bbox[10px,#ffa,border:1px groove navy]{7} \\ & \end{align}
Felix Marin
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AS a hint:an example
$$\int_{2}^{7} t^2 \, d\lfloor t \rfloor = 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 135$$ For the former, $t↦⌊t⌋$ increases only at its jumps on [$2,7]$, which are precisely $t=3,4,5,6,7$. Since the jump sizes are identically 1,
Khosrotash
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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 $(xi,x_{i+1})$ and the contribution to the integral will be $f(x∗i)(g(x_{i+1})−g(xi))$, where $x∗i \in (xi,x_{i+1})$. 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. – Khosrotash Jul 12 '20 at 18:17
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