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\( \int_{0}^{1} \frac{1}{\left\lfloor 1-\log _{2}(1-x)\right\rfloor} d x \) |
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Answer» \(\int\limits_0^1\frac1{[1-log_2(1-x)]}dx\) = \(\int\limits_0^1\cfrac1{[log_2\frac2{(1-x)}]}dx\) (\(\because log_22=1 \& log\,A- log\,B = log\frac{A}{B}\)) \(\therefore\) 0 < x < 1 ⇒ 0 < 1 - x < 1 ⇒ 2/(1-x) > 2 Now, \(\int\limits_0^1\cfrac1{[log_2\frac2{1-x}]}dx\) = ....+ \(\int\limits_{15/16}^{7/8}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{7/8}^{3/4}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{3/4}^{1/2}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{1/2}^{1}\cfrac1{[log_2\frac2{1-x}]}dx\) = .... + \(\int\limits_{15/16}^{7/8}\frac1{4}dx\) + \(\int\limits_{7/8}^{3/4}\frac1{3}dx\) + \(\int\limits_{3/4}^{1/2}\cfrac1{2}dx\) = ....+ \(\frac14(\frac{14}{16}-\frac{15}{16})+\frac13(\frac68-\frac78)\) + \(\frac12(\frac24-\frac34)+\frac12(1-\frac12)\) = \(\frac14-\frac18-\frac1{24}-\frac1{64}-..... <4\) Hence, \(\int\limits_0^1\frac1{1-log_2(1-x)}dx<1/4\) |
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