Partition this integral!

Calculus Level 2

∫ 1 ∞ 2 x { x } − { x } 2 x 2 ⌊ x ⌋ 2 d x \large \int_{1}^{\infty} \dfrac {2x\{x\} - \{x\}^2} {x^2 \lfloor x \rfloor ^2} \, dx

If the integral above can be expressed as π a − b c \dfrac{\pi^a - b} c , where a , b , c a,b,c are all positive integers, find a + b + c a+b+c .

Notations:


The answer is 14.

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Subh Mandal by subh mandal , 22, India

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6 solutions

Chew-Seong Cheong
Dec 11, 2016

Relevant wiki: Riemann Zeta Function

I = ∫ 1 ∞ 2 x { x } − { x } 2 x 2 ⌊ x ⌋ 2 d x Note that { x } = x − ⌊ x ⌋ = ∑ k = 1 ∞ ∫ k k + 1 2 x ( x − k ) − ( x − k ) 2 k 2 x 2 d x = ∑ k = 1 ∞ ∫ k k + 1 2 x 2 − 2 k x − x 2 + 2 k x − k 2 k 2 x 2 d x = ∑ k = 1 ∞ ∫ k k + 1 x 2 − k 2 k 2 x 2 d x = ∑ k = 1 ∞ ∫ k k + 1 ( 1 k 2 − 1 x 2 ) d x = ∑ k = 1 ∞ [ x k 2 + 1 x ] k k + 1 = ∑ k = 1 ∞ ( k + 1 − k k 2 + 1 k + 1 − 1 k ) = ∑ k = 1 ∞ ( 1 k 2 + 1 k + 1 − 1 k ) Note that ∑ k = 1 ∞ 1 k 2 = ζ ( 2 ) = π 2 6 = ζ ( 2 ) − 1 1 where ζ ( ⋅ ) denotes the Riemann zeta function. = π 2 6 − 1 = π 2 − 6 6 \begin{aligned} I & = \int_1^\infty \frac {2x \{x\} - \{x\}^2}{x^2\lfloor x \rfloor^2} dx & \small \color{#3D99F6} \text{Note that } \{x\} = x - \lfloor x \rfloor \\ & = \sum_{k=1}^\infty \int_k^{k+1} \frac {2x(x-k)-(x-k)^2}{k^2x^2} dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \frac {2x^2-2kx-x^2+2kx -k^2} {k^2x^2} dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \frac {x^2 - k^2} {k^2x^2} dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \left(\frac 1{k^2} - \frac 1{x^2} \right) dx \\ & = \sum_{k=1}^\infty \left[\frac x{k^2} + \frac 1x \right]_k^{k+1} \\ & = \sum_{k=1}^\infty \left(\frac {k+1-k}{k^2} + \frac 1{k+1} - \frac 1k \right)\\ & = \sum_{k=1}^\infty \left(\frac 1{k^2} + \frac 1{k+1} - \frac 1k \right) & \small \color{#3D99F6} \text{Note that }\sum_{k=1}^\infty \frac 1{k^2} = \zeta (2) = \frac {\pi^2}6 \\ & = \zeta(2) - \frac 11 & \small \color{#3D99F6} \text{where } \zeta(\cdot) \text{ denotes the Riemann zeta function.} \\ & = \frac {\pi^2}6 - 1 \\ & = \frac {\pi^2 - 6}6 \end{aligned}

⟹ a + b + c = 2 + 6 + 6 = 14 \implies a+b+c = 2+6+6 = \boxed{14}

16 Helpful 0 Interesting 6 Brilliant 0 Confused

I do not understand what you have done after the last summation.. somefunction(2) = pi square/6. Could you pls explain what that function is and how you got that sum as pi square/6?

Jaya Krishna - 3 years, 7 months ago

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I have added some explanation and a reference. Hope it is helpful. It is in fact the Basel problem that was first solved by Leonhard Euler .

Chew-Seong Cheong - 3 years, 7 months ago
Subh Mandal
Dec 10, 2016

20 Helpful 0 Interesting 1 Brilliant 0 Confused

how did u got integration of 1/[x]^2 from 0 to infinity ?

A Former Brilliant Member - 4 years, 4 months ago

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∫ 1 ∞ ⁣ ⁣ ⁣ 1 [ x ] 2 d x = ∑ n = 1 ∞ ∫ n n + 1 ⁣ ⁣ ⁣ 1 [ x ] 2 d x = ∑ n = 1 ∞ ∫ n n + 1 ⁣ ⁣ ⁣ 1 n 2 d x = ∑ n = 1 ∞ 1 n 2 = π 6 \displaystyle\intop^\infty_1\!\!\!\dfrac{1}{[x]^2}\mathrm{d}x=\displaystyle\sum_{n=1}^\infty\intop^{n+1}_n\!\!\!\dfrac{1}{[x]^2}\mathrm{d}x=\displaystyle\sum_{n=1}^\infty\intop^{n+1}_n\!\!\!\dfrac{1}{n^2}\mathrm{d}x=\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi}{6} .

Laurent Shorts - 4 years, 3 months ago

It's the value of reimann zeta function for s=2

Aanjaneya Pandey - 6 months, 2 weeks ago

nice one

incomplete steps

Shubham Rustagi - 4 years, 6 months ago

I did the same way.

Vraj Mistry - 4 years, 5 months ago
Anubhav Tyagi
Dec 20, 2016

2 x { x } − { x } 2 x 2 ⌊ x ⌋ 2 = { x } ( 2 x − { x } ) x 2 ⌊ x ⌋ 2 = { x } ( x + ( x − { x } ) ) x 2 ⌊ x ⌋ 2 = ( x − ⌊ x ⌋ ) ( x + ⌊ x ⌋ ) x 2 ⌊ x ⌋ 2 = x 2 − ⌊ x ⌋ 2 x 2 ⌊ x ⌋ 2 = 1 ⌊ x ⌋ 2 − 1 x 2 Hence we evaluate the integral as = ∫ 1 ∞ 1 ⌊ x ⌋ 2 d x − ∫ 1 ∞ 1 x 2 d x = ( ∫ 1 2 1 1 2 d x + ∫ 2 3 1 2 2 d x ⋯ ⋯ ) − ∫ 1 ∞ 1 x 2 d x = ( 1 1 2 + 1 2 2 + 1 3 2 ⋯ ⋯ ) − 1 = π 2 6 − 1 = π 2 − 6 6 \begin{aligned} &\dfrac {2x\{x\} - \{x\}^2} {x^2 \lfloor x \rfloor ^2}=\dfrac{\{x\}\big( 2x - \{x\}\big)} {x^2 \lfloor x \rfloor ^2}=\dfrac{\{x\}\big( x+\big(x - \{x\}\big)\big)} {x^2 \lfloor x \rfloor ^2}=\dfrac{\big(x- \lfloor x \rfloor\big)\big( x+ \lfloor x \rfloor\big)} {x^2 \lfloor x \rfloor ^2}=\dfrac{x^2- \lfloor x \rfloor ^2} {x^2 \lfloor x \rfloor ^2}=\frac{1}{\lfloor x \rfloor ^2}- \frac{1}{x^2}\\ &\text{Hence we evaluate the integral as }\\ &=\displaystyle \int_{1}^{\infty} \frac{1}{\lfloor x \rfloor ^2} \,dx - \displaystyle \int_{1}^{\infty} \frac{1}{x^2} \,dx \\ &= \bigg(\displaystyle \int_{1}^{2}\frac{1}{1^2} \,dx +\displaystyle \int_{2}^{3}\frac{1}{2^2}\,dx \cdots \cdots \bigg) - \displaystyle \int_{1}^{\infty} \frac{1}{x^2} \,dx \\ &=\bigg(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \cdots \cdots \bigg) - 1 \\ &= \frac{\pi ^2}{6} -1 = \frac{\pi^2-6}{6} \\ \end{aligned}

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Paul Hindess
Dec 10, 2016

Enjoyed this problem. Having never solved an integral involving either { x } \{x\} or [ x ] [x] , my approach is probably naive, but thought I would upload it for three reasons:

  1. as a Latex challenge for myself
  2. as a challenge to the reader (can anyone follow my thought process?!)
  3. to see if anyone else took a similar approach!

Here goes:

∫ 1 ∞ 2 x { x } − { x } 2 x 2 [ x ] 2 d x = ∑ n = 1 ∞ ∫ n n + 1 2 x { x } − { x } 2 x 2 [ x ] 2 d x = ∑ n = 1 ∞ ∫ 0 1 2 ( x + n ) x − x 2 ( x + n ) 2 n 2 d x = ∑ n = 1 ∞ 1 n 2 ∫ 0 1 x 2 + 2 n x x 2 + 2 n x + n 2 d x = ∑ n = 1 ∞ 1 n 2 ∫ 0 1 1 − n 2 x 2 + 2 n x + n 2 d x = ∑ n = 1 ∞ 1 n 2 ∫ 0 1 1 − n 2 ( x + n ) − 2 d x = ∑ n = 1 ∞ 1 n 2 [ x + n 2 x + n ] 0 1 = ∑ n = 1 ∞ 1 n 2 ( 1 + n 2 n + 1 − 0 − n 2 n ) = ∑ n = 1 ∞ 1 n 2 ( n + 1 ) \begin{aligned} \displaystyle \int_{1}^{\infty} \frac {2x\{x\} - \{x\}^2} {x^2[x]^2} dx & = \sum_{n=1}^{\infty} \int_{n}^{n+1} \frac {2x\{x\} - \{x\}^2} {x^2[x]^2} dx \\ & = \sum_{n=1}^{\infty} \int_{0}^{1} \frac {2(x+n)x - x^2} {(x+n)^2n^2} dx \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2} \int_{0}^{1} \frac {x^2+2nx} {x^2 + 2nx + n^2} dx \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2} \int_{0}^{1} 1- \frac {n^2} {x^2 + 2nx + n^2} dx \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2} \int_{0}^{1} 1- n^2(x+n)^{-2} dx \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2} \Big[x + \frac {n^2} {x+n}\Big]_0^1 \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2} \Big(1 + \frac {n^2} {n+1} - 0 - \frac {n^2} {n} \Big) \\ & = \sum_{n=1}^{\infty} \frac {1}{n^2(n+1)} \end{aligned}

And at this point I resorted to Wolfram Alpha because I was in too much of a hurry to check if my approach had been error-free and successful.

This gave me:

π 2 − 6 6 \frac {\pi^2 - 6}{6}

So a + b + c = 2 + 6 + 6 = 14 a+b+c= 2+6 + 6 = 14 .

Thanks for reading and well done if you followed even some of it! If anyone can suggest a non-Wolfram approach to the final integral, I'd be grateful.

8 Helpful 0 Interesting 0 Brilliant 1 Confused

Very neat solution! You might enjoy reading this article .

But, can you justify the interchanging of the summation and the integral in the first? Hint: Fubini's theorem.

For the final sum, you can break it up via partial fraction decompositions , to get

∑ n = 1 ∞ 1 n 2 ( n + 1 ) = ∑ n = 1 ∞ ( 1 n 2 + 1 n + 1 − 1 n ) = ∑ n = 1 ∞ 1 n 2 ⏟ = ζ ( 2 ) = π 2 / 6 + ∑ n = 1 ∞ ( 1 n + 1 − 1 n ) ⏟ = − 1 = π 2 − 6 6 , \sum_{n=1}^\infty \dfrac1{n^2 (n+1)} = \sum_{n=1}^\infty \left( \dfrac1{n^2} + \dfrac1{n+1} - \dfrac1n \right) = \underbrace{ \sum_{n=1}^\infty \dfrac1{n^2} }_{=\zeta(2) = \pi^2/6}+ \underbrace{\sum_{n=1}^\infty \left( \dfrac1{n+1} - \dfrac1n \right) }_{=-1} = \dfrac{\pi^2-6}6 ,

where the latter sum can be expressed as a telescoping series , ∑ n = 1 ∞ ( a n + 1 − a n ) = lim ⁡ N → ∞ ∑ n = 1 N ( a n + 1 − a n ) = lim ⁡ N → ∞ ( a N + 1 − a 1 ) = 0 − 1 = − 1. \displaystyle \sum_{n=1}^\infty (a_{n+1} - a_n) = \lim_{N\to\infty} \sum_{n=1}^N (a_{n+1} - a_n) = \lim_{N\to\infty} (a_{N+1} - a_1) = 0-1 = -1 .

Pi Han Goh - 4 years, 6 months ago

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Thank you. Should have thought have partial fractions! (-:

Paul Hindess - 4 years, 6 months ago

Yeah at first i kept thinking how to do it, but suddenly it struck me to do so by creating a general term for nth integral and then applying summation.

Raunak Agrawal - 4 years, 3 months ago

(Oh just realized Pi Han made a similar remark, so ignore.)

If by "final integral" you actually mean "final summation", then observe that by Partial Fractions ,

1 n 2 ( n + 1 ) = 1 n 2 + 1 n + 1 − 1 n . \frac{ 1 } { n^2 ( n+1 ) } = \frac{ 1}{ n^2 } + \frac{1}{n+1} - \frac { 1 } { n }.

Hence by the telescoping approach, the sum is ( ∑ 1 n 2 ) − 1 \left ( \sum \frac{1}{n^2 } \right) - 1 .

Calvin Lin Staff - 4 years, 6 months ago

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Thanks! Very helpful. And, yes - I've changed it to read "final summation". (-:

Paul Hindess - 4 years, 6 months ago
Kushal Bose
Dec 12, 2016

∫ 1 ∞ 2 x { x } − { x } 2 x 2 [ x ] 2 d x \displaystyle \int_{1}^{\infty} \dfrac{2x\{x\}-\{x\}^2}{x^2 [x]^2 } dx

= ∫ 1 ∞ ( 2 x − { x } ) { x } x 2 [ x ] 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{(2x-\{x\}) \{x\}}{x^2 [x]^2} dx

= ∫ 1 ∞ ( x + x − { x } ) { x } x 2 [ x ] 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{(x+ x-\{x\}) \{x\}}{x^2 [x]^2} dx

= ∫ 1 ∞ ( x + [ x ] ) { x } x 2 [ x ] 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{(x+[x]) \{x\}}{x^2 [x]^2} dx

= ∫ 1 ∞ ( x + [ x ] ) ( x − [ x ] ) x 2 [ x ] 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{(x+[x])(x-[x])}{x^2 [x]^2} dx

= ∫ 1 ∞ x 2 − [ x ] 2 x 2 [ x ] 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{x^2-[x]^2}{x^2 [x]^2} dx

= ∫ 1 ∞ 1 [ x ] 2 − 1 x 2 d x =\displaystyle \int _{1}^{\infty} \dfrac{1}{[x]^2} - \dfrac{1}{x^2} dx

Then it is a easy form .First part will give π 2 6 \frac{\pi^2}{6} and second part will give − 1 -1

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Can you explain how the 1st part gives pi square by 6

Nithin S - 8 months, 1 week ago
Bostang Palaguna
Dec 27, 2020

The idea is: We don't want to deal with { x } \{x\} so we utilize : { x } = x − ⌊ x ⌋ \{x\} = x - \left \lfloor{x}\right \rfloor

by doing some algebra, we would get: ∫ 1 ∞ x 2 − ⌊ x ⌋ 2 x 2 ⌊ x ⌋ 2 \int_1^\infty \frac{x^2 - \left \lfloor{x}\right \rfloor ^2}{x^2 \left \lfloor{x}\right \rfloor ^2}

make it up into two integral:

∫ 1 ∞ 1 / ⌊ x ⌋ 2 d x \int_1^\infty 1/\left \lfloor{x}\right \rfloor ^2 dx and ∫ 1 ∞ 1 / x 2 d x \int_1^\infty 1/x^2 dx

the second part is easy.

for the first part, imagine it in terms of area, you would have blocks with height of 1, 1/4, 1/9, 1/16, 1/25, ... and width of 1.

This is famously equal to π 2 / 6 \pi ^2 /6 (read : Riemann zeta function)

thus the integral result will be π 2 / 6 − 1 \pi ^2 /6 -1 .

∴ a + b + c = 14 \therefore a + b + c = \boxed{14}

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