 never appears continuously in the integral (only in the floor function), the integral itself is equal to the following sum:

Second, we can write the interior of the sum as:

However, because of the floor function, the denominator is actually just 

https://en.wikipedia.org/wiki/Natural_logarithm#Series

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[Daily puzzle] Speed integrating

south_korea_ln

First thing to notice is that because $$x$$ never appears continuously in the integral (only in the floor function), the integral itself is equal to the following sum:

$$
\sum_{k=1}^{\infty} \frac{1}{\left\lfloor \sum_{n=1}^{k} \frac{ k(-1)^{t+1}}{n^k} \right\rfloor} 
$$

Second, we can write the interior of the sum as:

$$
(-1)^{k+1} \frac{1}{\left\lfloor k \sum_{n=1}^{k} \frac{1}{n^k} \right\rfloor} 
$$

However, because of the floor function, the denominator is actually just $$k$$.

Thus:


$$
\sum_{k=1}^{\infty} \frac{1}{\left\lfloor \sum_{n=1}^{k} \frac{ k(-1)^{t+1}}{n^k} \right\rfloor}  = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} = \ln 2
$$

(For the last step, see here: https://en.wikipedia.org/wiki/Natural_logarithm#Series)


