The integral $\int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}$, where [ . ] denotes the greatest integer function, is equal to

<p>$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}$, let ${1 \over x} = t$</p> <p>${{ - 1} \over {{x^2}}}dx = dt$</p> <p>$$ = \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} } $$</p> <p>$$ = \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} } $$</p> <p>$$ = {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$$</p> <p>$$ = \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $$</p> <p>$$ = \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} } $$</p> <p>$$ = - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$$</p> <p>$= 1 + 6\log {6 \over 7}$</p>