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Compute the integral: $$\int_1^{\infty}\frac{1}{\left\lfloor \sum_{n=1}^{\left\lfloor x \right\rfloor}\frac{\left\lfloor x \right\rfloor(-1)^{\left\lfloor x+1 \right\rfloor}}{n^{\left\lfloor x \right\rfloor}} \right\rfloor}dx$$

In the video of the integration bee, the answer says $\ln(2)$, however, it seems like there was a mistake, but I could also done something wrong. Observe that $$=\int_1^2\frac{dx}{\left\lfloor 1 \right\rfloor}+\int_2^3\frac{dx}{\left\lfloor -2(1+\frac{1}{2^2})\right\rfloor}+\int_3^4\frac{dx}{\left\lfloor 3(1+\frac{1}{2^3}+\frac{1}{3^3}) \right\rfloor}+\int_4^5\frac{dx}{\left\lfloor -4(1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}) \right\rfloor}+...$$ $$\int_1^2 1dx+\int_2^3\frac{1}{-3}dx+\int_3^4\frac{1}{3}dx+\int_4^5\frac{1}{-5}dx+\int_5^6\frac{1}{5}dx+...=1???$$ I don't know, there is something off about this integral and I don't think the later terms follows the pattern...

Silver
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    Maybe $\int_1^\infty \frac{(-1)^{\lfloor x \rfloor +1}}{\left[\sum_{n=1}^{\lfloor x \rfloor} \frac{\lfloor x \rfloor (-1)^{n + 1}}{n^{\lfloor x \rfloor}}\right]} , dx $ ? – Ert33 Dec 08 '24 at 19:49
  • This does seem like a mistake in the solution because they may have gone the wrong way with the floor on the negative terms. – Ninad Munshi Dec 08 '24 at 20:28
  • @NinadMunshi that's what I think too – Silver Dec 08 '24 at 20:36
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    Fun adjustment, if the numerator becomes $i^{\lfloor x \rfloor}$, then the answer is $1-\frac{\pi}{4} + i\frac{\pi}{4}$. – Ninad Munshi Dec 08 '24 at 23:09
  • You can rewrite the integral as the telescoping sum $$\lim_{k\to\infty}\left(\int_{1}^{2}1dx+\int_{2}^{3}\frac{-1}{3}dx+\int_{3}^{4}\frac{1}{3}dx+\int_{4}^{5}\frac{-1}{5}dx+\int_{5}^{6}\frac{1}{5}dx,+\ldots+\int_{2k}^{2k+1}\frac{-1}{2k+1}dx+\int_{2k+1}^{2k+2}\frac{1}{2k+1}dx\right),.$$ Since all the terms cancel except for the first one, you end up with $$\lim_{k\to\infty}\int_{1}^{2}1dx = 1,.$$ I'm not sure where they got $\ln(2)$ from. :P – Accelerator Dec 09 '24 at 04:28

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