Evaluate $ \displaystyle \int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$.
(I am relatively confident of my answer below but I would still like to make sure that it is correct.)
Value of the integral $\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$
3
$\begingroup$
integration
improper-integrals
floor-function
1 Answers
2
\begin{align} \int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx &= \sum_{n=1}^\infty \int_n^{n+1} \frac n{x^3} \, dx \\ &= -\frac 12 \sum_{n=1}^\infty \frac n{x^2} \bigg\vert_n^{n+1} \\ &= \frac 12 \sum_{n=1}^\infty \left( \frac 1n - \frac n{(n+1)^2} \right) \\ &= \frac 12 \sum_{n=1}^\infty \frac{2n+1}{n(n+1)^2} \\ &= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \frac 1{n(n+1)^2} \\ &= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1} \right) - \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} \\ &= \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} + \frac 12 \\ &= \frac 12\sum_{k=2}^\infty \frac 1{k^2}+\frac 12 \\ &= \frac 12\left(\frac{\pi^2}6-1 \right)+\frac 12 \\ &= \boxed{\frac{\pi^2}{12}} \end{align}
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0You have a mistake in the second-to-last line. $k$ should range from $2$ to $\infty$. (Indeed, otherwise (1) you have a division by zero; (2) the change of indices is objectively incorrect; and, more subjective, (3) you get that extra $\frac{1}{2}$ in the result that "looks" uncanny.) Besides that, the rest (what is before) looks fine. – 2017-01-20
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0And now you have a mistake in the last line, after your edit. The $\frac{1}{2}$'s cancel. – 2017-01-20
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0It looks correct to me, now. (Conditioned on my not having overlooked anything.) – 2017-01-20