Integration of fractional part and floor function

Calculus Level 4

Evaluate the following definite integral :

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

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

Note: Use the following definition to solve the problem.

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

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

Put your answer to 3 decimal places.


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