$\int_0^1\frac{1}{7^{[1/x]}}dx$ $$\int_0^1\frac{1}{7^{[1/x]}}dx$$
Where $[x]$ is the floor function
now as the exponent is always natural, i converted it to an infinite sum
$$\sum\limits_{k=1}^{\infty} \frac{1}{7^{[1/k]}}$$
Which is an infinite GP, whose sum  is 6.
However, this is wrong, why is this wrong?

Note that when $x\in\left(\frac{1}{n+1},\frac{1}{n}\right],$ for some $n\in\mathbb{N},$ we have that $\lfloor\frac{1}{x}\rfloor=n.$ This observation motivates me to split up the given integral as such:
$$\begin{align*}
\int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\int_\frac{1}{2}^1\frac{dx}{7}+\int_\frac{1}{3}^\frac{1}{2}\frac{dx}{7^2}+\int_\frac{1}{4}^\frac{1}{3}\frac{dx}{7^3}+\ldots\\
&=\left(\frac{1}{7}+\frac{1}{2\times7^2}+\frac{1}{3\times7^3}+\ldots\right)-\left(\frac{1}{2\times7}+\frac{1}{3\times7^2}+\frac{1}{4\times7^3}+\ldots\right)\\
&=\boxed{1+6\ln{\left(\frac{6}{7}\right)}}
\end{align*}$$
I used the series expansions for $\ln(1+x)$ and $\ln(1-x)$ in the last step.
Note: Just so there is no confusion, I have more explicitly performed the calculation. From the observation, we get that
$$\begin{align*}
\int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\sum_{i\geq1}\int_\frac{1}{i+1}^\frac{1}{i}\frac{dx}{7^i}\\
&=\sum_{i\geq1}\left(\frac{1}{i\times7^i}-\frac{1}{(i+1)\times7^i}\right)\\
&=\sum_{i\geq1}\frac{1}{i\times7^i}-\sum_{i\geq1}\frac{1}{(i+1)\times7^i}\\
&=-\ln{\left(1-\frac{1}{7}\right)}-7\left(-\ln{\left(1-\frac{1}{7}\right)-\frac{1}{7}}\right)\\
&=\boxed{1+6\ln{\left(\frac{6}{7}\right)}}
\end{align*}$$
