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[Daily puzzle] Speed integrating
264 sats \ 1 comment \ @south_korea_ln 9 Dec 2024 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, 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: #797869 (solution in #798064).
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5000 sats \ 0 replies \ @SimpleStacker 13 Dec 2024
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)
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