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$$\int_0^\infty \frac{\sin(x\pi)}{\lceil x \rceil^2 + \lceil x \rceil} dx$$

My teacher recently gave me this and it's stumped me.

  • Do we have to find a closed form, or just study the convergence? The latter is immediate – Federico Dec 04 '18 at 20:14
  • He just said find the value of that. –  Dec 04 '18 at 20:17
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    I would suggest to take things slowly. Starting with $\lim_{n\to \infty} \int_0^n \frac{1}{\lceil x^2 \rceil + \lceil x \rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions. – Zacky Dec 04 '18 at 20:19
  • start dividing the integral in pieces where $\frac1{\lceil x^2\rceil +\lceil x\rceil}$ is constant – Masacroso Dec 04 '18 at 20:25
  • This is unpleasant, since it depends on the interleaving between the values of $n$ and $\sqrt{m}$ for $n,m\in\mathbb{N}$. Are you sure it is $\lceil x^2\rceil$ and not $\lceil x \rceil^2$? – Jack D'Aurizio Dec 04 '18 at 20:43
  • My bad yes it is ⌈x⌉^2 –  Dec 04 '18 at 21:22
  • @SamConnell heh, that changes the meaning of the question a lot! – YiFan Tey Dec 04 '18 at 23:00

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\sin\pars{\pi x} \over \left\lceil{x}\right\rceil^{2} + \left\lceil{x}\right\rceil} \,\dd x} = \sum_{k = 0}^{\infty}\int_{k}^{k + 1} {\sin\pars{\pi x} \over \pars{k + 1}^{2} + \pars{k + 1}}\,\dd x \\[5mm] = &\ {2 \over \pi}\sum_{k = 0}^{\infty} {\pars{-1}^{k} \over \pars{k + 1}\pars{k + 2}} = {2 \over \pi}\bracks{% \sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 1} - \sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 2}} \\[5mm] = &\ {2 \over \pi}\bracks{% -\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k} - \sum_{k = 2}^{\infty}{\pars{-1}^{k} \over k}} \\[5mm] = &\ -\,{2 \over \pi}\braces{% \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k} + \bracks{1 + \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k}}} \\[5mm] = &\ -\,{2 \over \pi}\bracks{% 1 + 2\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k}} = \bbx{4\ln\pars{2} - 2 \over \pi} \approx 0.2459 \end{align}

Felix Marin
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Hint: break the integral up into $$\int_0^\infty=\int_0^1+\int_1^2+\int_2^3+\cdots.$$ On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.

YiFan Tey
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