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)




SimpleStacker

science

Can you convince me this integral equals $$\ln(2)$$?
When I try to solve it, it equals $$1$$, but my reasoning is likely flawed.
$$\ln(2)$$ should be correct though, [as shown here](https://www.youtube.com/live/TykDs1CHLHk?feature=shared&t=3222), where the exercise comes from.

$$
\int_{1}^{\infty} \frac{1}{\left\lfloor \sum_{n=1}^{\lfloor x \rfloor} \frac{\lfloor x \rfloor (-1)^{\lfloor x + 1 \rfloor}}{n^{\lfloor x \rfloor}} \right\rfloor} \, dx
$$

In this competition, they were supposed to answer this within 4 minutes, but I'll leave the bounty open for as long as it takes :)

Previous iteration: https://stacker.news/items/797869/r/south_korea_ln (solution in https://stacker.news/items/797869/r/south_korea_ln?commentId=798064).


