$\int_0^1\frac{1}{7^{[1/x]}}dx$ $$\int_0^1\frac{1}{7^{[1/x]}}dx$$
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?
 A: 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
$$\begin{align*}
\int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\sum_{i\geq1}\int_\frac{1}{i+1}^\frac{1}{i}\frac{dx}{7^i}\\
&=\sum_{i\geq1}\left(\frac{1}{i\times7^i}-\frac{1}{(i+1)\times7^i}\right)\\
&=\sum_{i\geq1}\frac{1}{i\times7^i}-\sum_{i\geq1}\frac{1}{(i+1)\times7^i}\\
&=-\ln{\left(1-\frac{1}{7}\right)}-7\left(-\ln{\left(1-\frac{1}{7}\right)-\frac{1}{7}}\right)\\
&=\boxed{1+6\ln{\left(\frac{6}{7}\right)}}
\end{align*}$$
A: More generally,
$\begin{array}\\
\int_0^1 f([1/x])dx
&=-\int_1^{\infty} f([y])dy/y^2
\qquad y=1/x, x=1/y, dx=-1/y^2\\
&-=\sum_{n=1}^{\infty}\int_n^{n+1} \dfrac{f([y])dy}{y^2}\\
&=-\sum_{n=1}^{\infty}\int_n^{n+1} \dfrac{f([y])dy}{y^2}\\
&=-\sum_{n=1}^{\infty}f(n)\int_n^{n+1} \dfrac{dy}{y^2}\\
&=-\sum_{n=1}^{\infty}f(n)\dfrac1{y}\mid_n^{n+1} \\
&=-\sum_{n=1}^{\infty}f(n)(\dfrac1{n+1}-\dfrac1{n})\\
&=\sum_{n=1}^{\infty}f(n)(\dfrac1{n}-\dfrac1{n+1})\\
&=\sum_{n=1}^{\infty}\dfrac{f(n)}{n(n+1)}\\
\end{array}
$
If
$f(x)
=\dfrac1{c^x},
c > 1,
$
$\begin{array}\\
\int_0^1 f([1/x])dx
&=\sum_{n=1}^{\infty}f(n)(\dfrac1{n}-\dfrac1{n+1})\\
&=\sum_{n=1}^{\infty}\dfrac1{c^n}(\dfrac1{n}-\dfrac1{n+1})\\
&=\sum_{n=1}^{\infty}\dfrac1{nc^n}-\sum_{n=1}^{\infty}\dfrac1{(n+1)c^n}\\
&=\sum_{n=1}^{\infty}\dfrac1{c^n}\dfrac1{n}-c\sum_{n=1}^{\infty}\dfrac1{(n+1)c^{n+1}}\\
&=\sum_{n=1}^{\infty}\dfrac1{nc^n}-c\sum_{n=2}^{\infty}\dfrac1{nc^{n}}\\
&=\sum_{n=1}^{\infty}\dfrac1{nc^n}-c\left(\sum_{n=1}^{\infty}\dfrac1{nc^{n}}-\dfrac1{c}\right)\\
&=(1-c)\sum_{n=1}^{\infty}\dfrac1{nc^n}+1\\
&=-(1-c)\ln(1-\frac1{c})+1\\
&=(c-1)\ln(1-\frac1{c})+1\\
\end{array}
$
If $c=7$
this is
$1+6\ln(\frac67)
$.
