The integral ∫ _ 0 ^ 1 1 7 ^ [(1)/(x)] dx , where [·] denotes the...

$I = {∫}_{0}^{1}\dfrac{dx}{{7}^{\left[\dfrac{1}{x}\right]}} = {∫}_{0}^{1}{\left(\dfrac{1}{7}\right)}^{\left[\dfrac{1}{x}\right]}dx$

Let $\dfrac{1}{7} = k ⇒ I = {∫}_{0}^{1}{k}^{\left[\dfrac{1}{x}\right]}dx$

$I = {∫}_{0}^{1}{k}^{\left[\dfrac{1}{x}\right]}dx$

$⇒ I = {∫}_{\dfrac{1}{2}}^{1}{k}^{\left[\dfrac{1}{x}\right]}dx + {∫}_{\dfrac{1}{3}}^{\dfrac{1}{2}}{k}^{\left[\dfrac{1}{x}\right]}dx + {∫}_{\dfrac{1}{4}}^{\dfrac{1}{3}}{k}^{\left[\dfrac{1}{x}\right]}dx + ⋯$

$⇒ I = k\left(1 - \dfrac{1}{2}\right) + {k}^{2}\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + {k}^{3}\left(\dfrac{1}{3} - \dfrac{1}{4}\right) + ⋯$

$⇒ I = \left(k + \dfrac{{k}^{2}}{2} + \dfrac{{k}^{3}}{3} + ⋯\right) - \dfrac{1}{k}\left(\dfrac{{k}^{2}}{2} + \dfrac{{k}^{3}}{3} + ⋯\right)$

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

$⇒ I = - ln\dfrac{6}{7} - 7\left[- ln\dfrac{6}{7} - \dfrac{1}{7}\right]\left(∵ k = \dfrac{1}{7}\right)$

$∴ I = 1 + 6ln\dfrac{6}{7}$