Integration of fractional part and floor function

Calculus Level 5

Evaluate the following definite integral :

0 1 { ( 1 ) 1 x 1 x } d x \huge\int\limits_{0}^{1}\left\{ (-1)^{\left\lfloor \frac{1}{x} \right\rfloor} \frac{1}{x} \right\}\,\mathrm dx

where { x } \{x\} denotes the fractional part of x x and x \left\lfloor x\right\rfloor denotes the greatest integer function ( floor function ).

Note: Use the following definition to solve the problem.

{ x } = x x x R \left\{ x\right\} = x - \left\lfloor x\right\rfloor~\forall~x\in\Bbb R

where R \Bbb R denotes the set of all reals.

Put your answer to 3 decimal places.


The answer is 0.548.

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1 solution

Refaat M. Sayed
Sep 2, 2015

According to the given identity we can rewrite the internation in this form I = 0 1 { ( 1 ) 1 x 1 x } d x = 0 1 ( 1 ) 1 x 1 x d x 0 1 ( 1 ) 1 x 1 x d x I =\displaystyle\int_{0}^{1} \bigg\{ (-1)^{\left\lfloor \frac{1}{x} \right\rfloor} \frac{1}{x} \bigg\} dx = \displaystyle\int_{0}^{1} (-1)^{\left\lfloor \frac{1}{x} \right\rfloor} \frac{1}{x} dx - \displaystyle \int \limits^{1}_{0}\left\lfloor \left( -1\right) ^{\left\lfloor \frac{1}{x} \right\rfloor }\frac{1}{x} \right\rfloor dx Now let 1 x = y \frac{1}{x} = y . We get the internation above as I = 1 ( 1 ) y 1 y d y 1 ( 1 ) y y 1 y 2 d y I =\displaystyle\int \limits^{\infty }_{1}\left( -1\right) ^{\left\lfloor y\right\rfloor }\frac{1}{y} dy - \displaystyle\int \limits^{\infty }_{1}\left\lfloor \left( -1\right) ^{\left\lfloor y\right\rfloor }y\right\rfloor \frac{1}{y^{2}} dy Now I 1 = 1 ( 1 ) y 1 y d y = m = 1 ( 1 ) m m m + 1 d y y = m = 1 ( 1 ) m l n ( m + 1 m ) I_{1} = \displaystyle\int \limits^{\infty }_{1}\left( -1\right) ^{\left\lfloor y\right\rfloor }\frac{1}{y} dy = \large\sum \limits^{\infty }_{m=1}\left( -1\right) ^{m}\displaystyle\int \limits^{m+1}_{m}\frac{dy}{y} = \large\sum \limits^{\infty }_{m=1}\left( -1\right) ^{m} ln\bigg( \frac{m+1}{m} \bigg) so I 1 = l n [ m = 1 ( 2 m 2 m 1 × 2 m 2 m + 1 ) ] = l n ( π 2 ) I_{1} = -ln\bigg[\large\prod \limits^{\infty }_{m =1}\left( \frac{2m}{2m -1} \times \frac{2m }{2m +1} \right) \bigg] = -ln\bigg(\frac{\pi}{2} \bigg) I 2 = 1 ( 1 ) y y 1 y 2 d y = m = 1 m m + 1 ( 1 ) m y 1 y 2 d y I_{2} =\displaystyle\int \limits^{\infty }_{1}\left\lfloor \left( -1\right) ^{\left\lfloor y\right\rfloor }y\right\rfloor \frac{1}{y^{2}} dy =\large\sum \limits^{\infty }_{m=1}\displaystyle \int \limits^{m+1}_{m}\left\lfloor \left( -1\right) ^{m}y\right\rfloor \frac{1}{y^{2}} dy then I 2 = m = 1 2 m ( 2 m 2 m + 1 1 y 2 d y 2 m 1 2 m 1 y 2 d y ) I_{2} = \large\sum \limits^{\infty }_{m =1}2m \left( \int \limits^{2m +1}_{2m}\frac{1}{y^{2}} dy - \int \limits^{2m}_{2m -1}\frac{1}{y^{2}} dy\right) so I 2 = m = 1 1 2 m + 1 m = 1 1 2 m 1 = 1 I_{2}= \large\sum \limits^{\infty }_{m=1}\frac{1}{2m +1} - \large\sum \limits^{\infty }_{m=1}\frac{1}{2m -1} = -1 And : I = I 1 I 2 = 1 l n ( π 2 ) = 0.548 \text{And} : I = I_{1}- I_{2} = 1- ln\bigg(\frac{\pi}{2}\bigg) = \boxed {0.548}

@Refaat M. Sayed Nice work! But when I computed the integral numerically, I was getting an answer of 0.1447 -0.1447 ??

Kunal Gupta - 5 years, 8 months ago

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Could you tell me How you did that? .

Refaat M. Sayed - 5 years, 8 months ago

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@Refaat M. Sayed How do I add an image?? in comments

Kunal Gupta - 5 years, 8 months ago

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@Kunal Gupta @Hasan Kassim say that :first upload any pic u want to insert, on www.postimage.org, they provide a link for that image.

while writing solutions , write the following:

![](link

You should replace the word "link" by the link provided by that website, then close the paranthesis... and I think Master. @Calvin Lin will help us

Refaat M. Sayed - 5 years, 8 months ago

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@Refaat M. Sayed @Refaat M. Sayed thanks!! here you go go

Kunal Gupta - 5 years, 8 months ago

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@Kunal Gupta But the correct graph like this http://s11.postimg.org/lscirf7ur/Screenshot.png

Refaat M. Sayed - 5 years, 7 months ago

Really nice way of computing that logarithm sum! Innovative! I really liked it. I have always used polylogarithm or zeta function derivative for such sums but this wake me up. Thanks!

Kartik Sharma - 5 years, 8 months ago

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Thanks Kartik :)

Refaat M. Sayed - 5 years, 8 months ago

How is that pi/2?

Radha Koutha - 4 years, 9 months ago

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this product is well known as Wallis product and here is the proof

Refaat M. Sayed - 4 years, 9 months ago

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