A simple double integral.

Calculus Level 5

Let : $I= \int_0^1 \int_0^1 \left\{ \frac{x}{y} \right\} \ \mathrm{d}x\ \mathrm{d}y$ Find $\lfloor1000I\rfloor$ .

$\{x\}$ denotes to the fractional part of $x$ , and $\lfloor x\rfloor$ denotes to the floor function for $x$ .

Remark : Numerical integration using software can be wrong.

The answer is 461.

2 solutions

Pranav Arora
Jul 15, 2014

For the integral with respect to $x$ , use the substitution $\dfrac{x}{y}=t$ .

$\displaystyle I=\int_0^1 \int_0^{1/y} y\,\{t\}\,dt\,dy =\int_0^1 \int_0^1 y\,\{t\}\,dt\,dy+\int_1^{\infty} \int_0^{1/t} y\,\{t\}\,dy\,dt$

$\displaystyle \Rightarrow I=I_1+I_2$

$I_1$ is easy to evaluate and is equal to $\dfrac{1}{4}$ .

$\displaystyle \begin{aligned} I_2=\frac{1}{2}\int_1^{\infty} \frac{\{t\}}{t^2}\,dt & =\lim_{n\rightarrow \infty} \frac{1}{2}\sum_{k=1}^n \int_k^{k+1} \frac{t-k}{t^2}\,dt \\ &= \lim_{n\rightarrow \infty} \frac{1}{2}\sum_{k=1}^n\left(\ln\left(1+\frac{1}{k}\right)-\frac{1}{k+1}\right) \\ &=\lim_{n\rightarrow \infty} \frac{1}{2}\left(\ln(n+1)-H_{n+1}+1\right) \\ &=\frac{1-\gamma}{2} \\ \end{aligned}$

Hence,

$\displaystyle I=\frac{1}{4}+\frac{1-\gamma}{2}=\boxed{\dfrac{3}{4}-\dfrac{\gamma}{2}}$

$H_{n+1}$ is $n+1^{th}$ harmonic number .
$\gamma$ is the Euler-Mascheroni Constant .

Could you please explain the first step of your solution Pranav?

I did it in a little differently:

$\displaystyle\int_{0}^{1}\bigg( \displaystyle\int_{0}^{y} \frac{x}{y}\text{ }\text{d}x+\displaystyle\int_{y}^{2y} \frac{x}{y}-1\text{ }\text{d}x+\displaystyle\int_{2y}^{3y} \frac{x}{y}-2\text{ }\text{d}x+...+\displaystyle\int_{(\left \lfloor{n}\right \rfloor -1)y}^{\left \lfloor{n}\right \rfloor y} \frac{x}{y}-(\left \lfloor{n}\right \rfloor -1)\text{ }\text{d}x+\displaystyle\int_{\left \lfloor{n}\right \rfloor y}^{1} \frac{x}{y}-\left \lfloor{n}\right \rfloor\text{ }\text{d}x\bigg)\text{d}y$

where $ny=1$

Integrating w.r.t. $x$ we obtain:

$\displaystyle\int_{0}^{1} \frac{\left \lfloor{n}\right \rfloor(\left \lfloor{n}\right \rfloor +1)y}{2}+\dfrac{1}{2y}-\left \lfloor{n}\right \rfloor\text{ }\text{d}y$

Now putting $n=\dfrac{1}{y}$ and $y=\dfrac{1}{t}$ we obtain:

$\displaystyle\int_{1}^{\infty} \frac{(\left \lfloor{t}\right \rfloor)^{2}+\left \lfloor{t}\right \rfloor}{2t^{3}}+\dfrac{1}{2t}-\frac{\left \lfloor{t}\right \rfloor}{t^{2}}\text{ }\text{d}t$

$=\displaystyle\sum_{r=1}^{\infty}\!\!\displaystyle\int_{r}^{r+1} \frac{r^{2}+r}{2t^{3}}+\frac{1}{2t}-\frac{r}{t^{2}}\text{ }\text{d}t$

$=\dfrac{1}{4}+\dfrac{1}{2}\displaystyle\sum_{r=1}^{\infty} \ln \left( \dfrac{r+1}{r}\right)-\dfrac{1}{r+1}$

$=\dfrac{1}{4}+\dfrac{1}{2}(1-\gamma)$

However I didn't know about this number $\gamma$ and had to evaluate the final limit on W|A.

Karthik Kannan - 6 years, 11 months ago

Your solution is nice too. :)

Are you asking about the step when I split the integral into two? If so, try graphing the region I am integrating on.