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

We need to evaluate $$I = \int_0^1 \frac{1}{7^{[1/x]}}\,dx$$ where $$[\cdot]$$ denotes the greatest integer function.

Let $$t = \frac{1}{x}$$, so $$x = \frac{1}{t}$$ and $$dx = -\frac{1}{t^2}\,dt$$. When $$x \to 0^+$$, $$t \to \infty$$; when $$x = 1$$, $$t = 1$$. So $$I = \int_{\infty}^{1} \frac{1}{7^{[t]}} \cdot \left(-\frac{1}{t^2}\right) dt = \int_1^{\infty} \frac{1}{7^{[t]} \cdot t^2}\,dt$$.

On the interval $$[n, n+1)$$, $$[t] = n$$, so we split: $$I = \sum_{n=1}^{\infty} \frac{1}{7^n} \int_n^{n+1} \frac{1}{t^2}\,dt = \sum_{n=1}^{\infty} \frac{1}{7^n} \left[-\frac{1}{t}\right]_n^{n+1} = \sum_{n=1}^{\infty} \frac{1}{7^n} \left(\frac{1}{n} - \frac{1}{n+1}\right).$$ This equals $$\sum_{n=1}^{\infty} \frac{1}{7^n} \cdot \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \frac{1}{7^n}\left(\frac{1}{n} - \frac{1}{n+1}\right).$$

Therefore $$I = \sum_{n=1}^{\infty} \frac{1}{n \cdot 7^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot 7^n}.$$ For the second sum, let $$m = n + 1$$: $$\sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot 7^n} = \sum_{m=2}^{\infty} \frac{1}{m \cdot 7^{m-1}} = 7\sum_{m=2}^{\infty} \frac{1}{m \cdot 7^m}.$$

Using the known series $$-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$$ for $$|x| < 1$$, we get $$\sum_{n=1}^{\infty} \frac{1}{n \cdot 7^n} = -\ln\left(1 - \frac{1}{7}\right) = -\ln\frac{6}{7} = \ln\frac{7}{6}.$$ Hence $$I = \sum_{n=1}^{\infty} \frac{1}{n \cdot 7^n} - 7 \sum_{m=2}^{\infty} \frac{1}{m \cdot 7^m} = \ln\frac{7}{6} - 7\left(\ln\frac{7}{6} - \frac{1}{7}\right) = 1 - 6\ln\frac{7}{6} = 1 + 6\ln\frac{6}{7}.$$

The answer is Option B: $$1 + 6\ln\left(\frac{6}{7}\right)$$.