Quite happy as a sum, why the ratio?

Calculus Level 5

$\displaystyle \int_{1}^{10} \dfrac{\{x\}}{\lfloor x \rfloor} \ dx$

If the value of definite integral above can be written as $\dfrac{a}{b}$ for coprime positive integers, find $a+b$ .

Details and assumptions :

Every $x\in \mathbb{R}$ can be written as $x=\lfloor x \rfloor + \{x\}$ .
$\lfloor x \rfloor$ denotes greatest integer less than or equal to $x$ .
$\{x\}$ is the fractional part of $x$ .

Also see product and sum of squares .

Read the brilliant.org wiki for further details.

The answer is 12169.

2 solutions

Aditya Raut
Mar 30, 2015

As per the definition of integral as Area under the curve ,

the graph of the function $\dfrac{\{x\}}{\lfloor{x}\rfloor}$ would be as follows, in the form of $9$ triangles

Bases of all these triangles are $1$ , but their heights are $\dfrac{1}{1},\dfrac{1}{2}, \dfrac{1}{3}, .... \dfrac{1}{9}$

Sum of their areas will be $\dfrac{1}{2} \biggl( 1+\dfrac{1}{2}+ \dfrac{1}{3}+ .... +\dfrac{1}{9} \biggr) = \dfrac{7129}{5040}$

Answer $7129+5040= \boxed{12169}$

I was too lazy to calculate the sum so i couldn't solve but here is my method, $\{x\}=x-[x]\\ \Longrightarrow \int_{1}^{10}\dfrac{\{x\}}{[x]}=\int_{1}^{10}=\dfrac{x-[x]}{[x]}=\int_{1}^{10}\dfrac{x}{[x]}dx-9\\ \int_{1}^{10}\dfrac{x}{[x]}dx=\int_{1}^{2}\dfrac{x}{1}+\int_{2}^{3}\dfrac{x}{2}+...\int_{9}^{10}\dfrac{x}{9} \\= \sum_{n=1}^{9}\dfrac{((n+1)^2-n^2)}{2n} \\ =\sum_{n=1}^9\dfrac{2n+1}{2n}\\ =\sum_{n=1}^9\dfrac{1}{2n}+9(this\ gets\ cancelled)\\ \Longrightarrow our\ answer\ is \dfrac{1}{2}(H_9)$

Adarsh Kumar - 5 years, 11 months ago

Aditya Raut If this question comes in your exam,how will you draw the graph of $\large \frac{\{x\}}{[x]}$ ?

Akhil Bansal - 5 years, 9 months ago

Shriram Lokhande
Mar 30, 2015

\[ \begin{array} {} \displaystyle \int_1^{10} {\dfrac{\{x\}}{\lfloor x \rfloor}dx} & = \displaystyle \sum_{k=1}^{9} {\int_{k}^{k+1} {\dfrac{\{x\}}{\lfloor x \rfloor}dx}} \\ & = \displaystyle \sum_{k=1}^{9} {\int_{k}^{k+1} {\dfrac{\{x\}}{k}dx}} \\ & = \displaystyle \sum_{k=1}^9 {\dfrac{1}{k} \int_0^1 {x dx}} = \dfrac{H_9}{2} \\ & = \dfrac{7129}{5040} \end{array} \]

where , $H_n$ is the n-th harmonic number.

Hence, a + b = $\boxed{12169}$

for H9 ,http://users.encs.concordia.ca/~chvatal/notes/harmonic.html

tapas pal - 6 years, 2 months ago

Yes, they have provided a list of the fractions till $H_9$ . But what if someone wants the fraction form of $H_{20}$ ?

I know the approximation method (which is actually very trivial) which provides tight enough bounds for $H_x$ using the integral of $f(x)=\dfrac{1}{x}$ but that doesn't give us an exact result in fraction form! I think the calculation of $H_9$ was meant to be bashy since as far as I know, there is no perfect way of getting the exact fraction value of $H_x$ for large enough $x$ . Only manual tedious fraction addition seems like the way to go!

Prasun Biswas - 6 years, 2 months ago

What I want to know is how did you calculate the fraction form of $H_9$ ? Any special methods or just adding fractions manually?