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I'm working on an integral problem(the rest of which is irrelevant) and this integral arises, which has stumped me. $$\int_{0}^{1}\int_{0}^{x}\left\lfloor\frac{1}{1-t}\right\rfloor dt \,dx$$

$\bf\lfloor\quad\rfloor$ is Floor function.

Looking at $\int\limits_{0}^{x}\left\lfloor\dfrac{1}{1-t}\right\rfloor dt$:

Clearly this diverges for x = 1, but is there any representation in terms of $\zeta(x)$?

So far I have:

$$\int_{0}^{1}\left\lfloor\frac{1}{1-t}\right\rfloor dt = \sum_{n=0}^{\infty}\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}} \thinspace (n+1)\,dt = \zeta(1)-1$$

But I am unsure how to address the issue of finding what interval x is between and calculate the area up to x in relation to x since there are different bounds x could be located in. I suppose something like this could work but it doesn't seem like something that could be further integrated form 0 to 1.

$$A = \left\lfloor\frac{1}{1-x}\right\rfloor-1$$ $$\sum_{n=0}^{A}\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}} \thinspace (n+1)\,dt + \int_{\frac{A}{A+1}}^{x}(A+1)\,dt$$

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  • $\begingroup$ Is there a dependency of x and t? $\endgroup$ – Moti May 17 '16 at 5:08
  • $\begingroup$ @Moti Well t is a u sub for xy but I didn't think that would be important. I have properly done the dt and limits. $\endgroup$ – Jasper Braun May 17 '16 at 11:19
  • $\begingroup$ @JasperBraun I just add \left and \right to the "floor's". $\endgroup$ – Felix Marin May 17 '16 at 20:12
  • $\begingroup$ It's odd that you stopped where you did, rather than compute all of the integrals. $\endgroup$ – Hurkyl May 17 '16 at 20:14
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Denote $f(t) = \lfloor 1/(1-t) \rfloor$, and consider the function $$ g(x,t) = \Theta(x - t) f(t), $$ where $\Theta(u)$ is the Heaviside step function. Since $g(x,t) = f(t)$ on the original region of integration, we can expand the region of integration to the unit square: $$ \int_{0}^{1}\int_{0}^{x} f(t) \, dt \, dx = \int_0^1 \int_0^1 \Theta(x - t) f(t) \, dt \, dx. $$ The function $g(x,t)$ is non-negative and measurable, so by Tonelli's Theorem (credit to @RRL for pointing this out in the comments), we can exchange the order of integration: $$ \int_0^1 \int_0^1 \Theta(x - t) f(t) \, dt \, dx = \int_0^1 \int_0^1 \Theta(x - t) f(t) \, dx \, dt = \int_0^1 \left[ \int_t^1 f(t) \, dx \right] dt. $$

This integral can then be performed: \begin{align*} \int_{0}^{1} \left[ \int_{t}^{1} f(t) \, dx \right] dt &= \int_0^1 (1 - t) f(t) \, dt \\ &= \sum_{n = 0}^\infty \int_{n/(n+1)}^{(n+1)/(n+2)} (1-t) (n+1) \, dt \\ &= \sum_{n = 0}^\infty (n+1)\left[ t - \frac{t^2}{2} \right]_{n/(n+1)}^{(n+1)/(n+2)} \\ &= \sum_{n = 0}^\infty \frac{3 + 2n}{2(n+1)(n+2)^2} \\ &= \sum_{n = 0}^\infty \left[ \frac{1}{2(n+1)} - \frac{1}{2(n+2)} + \frac{1}{2 (n+2)^2} \right] \\ &= \sum_{n = 1}^\infty \left[ \frac{1}{2n} - \frac{1}{2(n+1)}\right] + \frac{1}{2} \sum_{n=2}^\infty \frac{1}{n^2} \\&= \frac{1}{2} + \frac{1}{2} (\zeta(2) - 1) = \frac{\zeta(2)}{2} = \frac{\pi^2}{12}. \end{align*} In the last step, we have used telescoping series to perform the first sum, and the second sum is the definition of $\zeta(2)$ except that it's missing the $n = 1$ term ($1/1^2$).

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    $\begingroup$ Nice +1. Since $f$ is non-negative and measurable the interchange is fine(Tonelli's theorem). To skip the geometric argument -- $\int_0^1 \int_0^x f(t) \, dt \,dx = \int_0^1 \int_0^1 f(t)1_{\{t \leqslant x\}} \, dt \,dx = \int_0^1 \int_0^1 f(t)1_{\{x \geqslant t\}} \, dx \,dt = \int_0^1 \int_t^1 f(t) \, dx \,dt $. $\endgroup$ – RRL May 17 '16 at 21:30
  • $\begingroup$ @RRL: Thanks for the tips. I've edited the answer accordingly. $\endgroup$ – Michael Seifert May 17 '16 at 21:46
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{1}\int_{0}^{x}\left\lfloor{1 \over 1 - t}\right\rfloor\,\dd t\,\dd x & = \int_{0}^{1}\left\lfloor{1 \over 1 - t}\right\rfloor\,\int_{t}^{1}\dd x\,\dd t = \int_{0}^{1}\left\lfloor{1 \over 1 - t}\right\rfloor\pars{1 - t}\,\dd t = \int_{0}^{1}\left\lfloor{1 \over t}\right\rfloor t\,\dd t \\[3mm] &= \int_{1}^{\infty}{\left\lfloor t\right\rfloor \over t^{3}}\,\dd t = \int_{1}^{2}{1 \over t^{3}}\,\dd t + \int_{2}^{3}{2 \over t^{3}}\,\dd t + \cdots \\[3mm] & = \int_{1}^{2}\bracks{{1 \over t^{3}} + {2 \over \pars{t + 1}^{3}} + \cdots} \,\dd t = \int_{1}^{2}\sum_{k = 0}^{\infty}{k + 1\over \pars{t + k}^{3}} \,\dd t \\[3mm] & = \int_{1}^{2}\bracks{% \Psi\,'\pars{t} + \half\pars{t - 1}\Psi\,''\pars{t}}\,\dd t \\[3mm] & = \Psi\pars{2} - \Psi\pars{1} + \half\,\Psi\,'\pars{2} - \half\int_{1}^{2}\Psi\,'\pars{t}\,\dd t \\[3mm] & = \half\bracks{\Psi\pars{2} - \Psi\pars{1}} + \half\,\Psi\,'\pars{2} = \color{#f00}{\pi^{2} \over 12} \approx 0.8225 \end{align}

$\Psi$ is the digamma function.

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