QJEEM22Ma8046: JEE Main 2022 Mathematics - Class 12 JEEM22Ma8046 Solution: The integral , where [ . ] denotes the greatest integer…

$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}$ , let ${1 \over x} = t$

${{ - 1} \over {{x^2}}}dx = dt$

$= \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} }$

$= \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} }$

$= {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$

$= \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)}$

$= \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} }$

$= - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$

$= 1 + 6\log {6 \over 7}$