# Integral: $\int_{0}^{x}\lfloor\frac{1}{1-t}\rfloor \,dt$

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.

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

• Is there a dependency of x and t? – Moti May 17 '16 at 5:08
• @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. – Jasper Braun May 17 '16 at 11:19
• @JasperBraun I just add \left and \right to the "floor's". – Felix Marin May 17 '16 at 20:12
• It's odd that you stopped where you did, rather than compute all of the integrals. – Hurkyl May 17 '16 at 20:14

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 ($1/1^2$).
• Nice +1. Since $f$ is non-negative and measurable the interchange is fine(Tonelli's theorem). To skip the geometric argument -- $\int_0^1 \int_0^x f(t) \, dt \,dx = \int_0^1 \int_0^1 f(t)1_{\{t \leqslant x\}} \, dt \,dx = \int_0^1 \int_0^1 f(t)1_{\{x \geqslant t\}} \, dx \,dt = \int_0^1 \int_t^1 f(t) \, dx \,dt$. – RRL May 17 '16 at 21:30
$\Psi$ is the digamma function.