The integral ${\int }_0^1\frac{1}{7^{\left[\frac{1}{x}\right]}}dx$, where $\left[·\right]$ denotes the greatest integer function, is equal to

$I={\int }_0^1\frac{dx}{7^{\left[\frac{1}{x}\right]}}={\int }_0^1{\left(\frac{1}{7}\right)}^{\left[\frac{1}{x}\right]}dx$

Let $\frac{1}{7}=k \Rightarrow I={\int }_0^1{k}^{\left[\frac{1}{x}\right]}dx$

$I={\int }_0^1{k}^{\left[\frac{1}{x}\right]}dx$

$\Rightarrow I={\int }_{\frac{1}{2}}^1{k}^{\left[\frac{1}{x}\right]}dx+{\int }_{\frac{1}{3}}^{\frac{1}{2}}{k}^{\left[\frac{1}{x}\right]}dx+{\int }_{\frac{1}{4}}^{\frac{1}{3}}{k}^{\left[\frac{1}{x}\right]}dx+\cdots$

$\Rightarrow I=k\left(1-\frac{1}{2}\right)+k^2\left(\frac{1}{2}-\frac{1}{3}\right)+k^3\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots$

$\Rightarrow I=\left(k+\frac{k^2}{2}+\frac{k^3}{3}+\cdots \right)-\frac{1}{k}\left(\frac{k^2}{2}+\frac{k^3}{3}+\cdots \right)$

$\Rightarrow I=-\ln \left(1-k\right)-\frac{1}{k}\left[-\ln \left(1-k\right)-k\right]$

$\Rightarrow I=-\ln \frac{6}{7}-7\left[-\ln \frac{6}{7}-\frac{1}{7}\right] \left(\because k=\frac{1}{7}\right)$

$\therefore I=1+6\ln \frac{6}{7}$