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\(\displaystyle%20\int_{0}^{1}\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}dx\)

where \(\left\{x\right\} =\) fractional part function

Boa tarde.
Aconselho que visite o seguinte site: http://mathworld.wolfram.com/FractionalPart.html
Obrigado.

Hi kinu

just some tips...

You have to split it in parts

the \(x=1/2\) is an important point

you split in parts the integral on the set \(x\in[\frac{1}{2},1]\)

\(\displaystyle%20\int_{0}^{1}\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}dx\)

if
\(f(x)=\left\{\frac{1}{x}\right\}^2.\left\{\frac{1}{1-x}\right\}\)

Then

\(\int_{1/2}^{2/3}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-2\right)dx\)

\(\int_{2/3}^{3/4}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-3\right)dx\)

\(\int_{3/4}^{4/5}f(x)dx=\int_{1/2}^{2/3}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-4\right)dx\)

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.
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\(\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}}f(x)dx=\int_{\frac{n}{n+1}}^{\frac{n+1}{n+2}}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-(n+1)\right)dx\)




Then you do it for the second set \(x\in[0,\frac{1}{2}]\)

\(\int_{1/3}^{1/2}f(x)dx=\int_{1/3}^{1/2}\left(\frac{1}{x}-1\right)^2\left(\frac{1}{1-x}-1\right)dx\)

\(\int_{1/4}^{1/3}f(x)dx=\int_{1/4}^{1/3}\left(\frac{1}{x}-2\right)^2\left(\frac{1}{1-x}-1\right)dx\)

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\(\int_{\frac{1}{n}}^{\frac{1}{n-1}}f(x)dx=\int_{\frac{1}{n}}^{\frac{1}{n-1}}\left(\frac{1}{x}-(n-2)\right)^2\left(\frac{1}{1-x}-1\right)dx\)

Then calculate the finite integrals and at the end sum everything, you should get two infinite series...

Take care.

PS: kinu, after 57 posts with very dificult exercises to solve, your contributions to others have been ZERO
We have no commitments here, and our target is to help other people on a philanthropic basis, but we have no ads, we don't want donations, and we solve problems for free, our core are our great math contributors, and at this stage you have posted 57 problems very difficult to solve and you have solved ZERO exercises.
Do not destroy the spirit!!!!

_________________
João Pimentel Ferreira
 
_{Partilhe dúvidas e resultados, ajude a comunidade com a sua pergunta!
Não lhe dês o peixe, ensina-o a pescar (provérbio chinês)
Fortalecemos a quem ajudamos pouco, mas prejudicamos se ajudarmos muito (pensamento budista)}

Thanks João P. Ferreira for your Hint

and extremelly Soory from my side.

These types of things not happen in future

Thanks