The integral _0^117^[1/x],dx, where [] denotes the greatest integer function, is equal to

Question

The integral $\displaystyle\int_0^1\frac1{7^{[1/x]}}\,dx$, where $[\cdot]$ denotes the greatest integer function, is equal to

Accepted Answer

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

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

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

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

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

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

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

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

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

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