How do I integrate $$\int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx$$
Where $\lceil x \rceil $ is the ceiling function, and $\left\{x\right\}$ is the fractional part function
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9I don't understand why you changed the question completely. It might be an idea to reverse it and pose this as a new question. – Dec 27 '12 at 21:24
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3I also don't understand that part of "simple integral": if it is "simple" then what's the problem? – DonAntonio Dec 27 '12 at 23:13
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2If I write simple it attracts more viewers and for all I know it is simple to people familiar with this kind of thing. – Ethan Splaver Dec 28 '12 at 10:19
3 Answers
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Hint: try to use the definition of the fractional part function which is defined by
$$ \left\{ x\right\} = x - \lfloor x\rfloor , $$
and the following relation between the floor and ceiling functions
$$ \lceil x \rceil - \lfloor x \rfloor = \begin{cases} 0&\mbox{ if } x\in \mathbb{Z}\\ 1&\mbox{ if } x\not\in \mathbb{Z} \end{cases}. $$
Added:
$$ \int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx=\int_{0}^1 x (1+\lfloor 1/x \rfloor)(1/x-\lfloor1/x\rfloor)\, dx. $$
Now, make the change of variables $y=1/x$ to the last integral
$$\int_{0}^1 x (1+\lfloor 1/x \rfloor)(1/x-\lfloor1/x\rfloor)\, dx=\int_{1}^{\infty} \frac{1}{y} (1+\lfloor y \rfloor)(y-\lfloor y\rfloor)\, \frac{dy}{y^2}$$
$$\implies I = \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{1}{y^3} (1+n)(y-n)\, dy= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{2} $$
Note: To evaluate the sum, use the telescoping technique. First write the summand as
$$ \frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}. $$
Now, find the partial sum of the series
$$ s_n = \sum_{k=1}^{n} \left( \frac{1}{k}-\frac{1}{k+1} \right)=1-\frac{1}{n+1}. $$
Then the series sums to
$$ s = \lim_{n \to \infty} s_n = 1. $$
Mhenni Benghorbal
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Ive been trying for the past 30 minutes I really dont understand, I really need to brush up on my integration skills, I would appreciate it if you walked me through it. – Ethan Splaver Dec 27 '12 at 21:32
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The main idea is to divide $(0,1)$ into "good" intervals. I'lll give only the main steps of computation $$ \int\limits_{(0,1)} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx =\sum\limits_{n=1}^\infty\int\limits_{n\leq \frac{1}{x}<n+1} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx =\sum\limits_{n=1}^\infty\int\limits_{\frac{1}{n+1}< x\leq \frac{1}{n}} x (n+1) \left(\frac{1}{x}-n\right) dx =\sum\limits_{n=1}^\infty\frac{1}{2n^2+2n}=\frac{1}{2}\sum\limits_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2} $$
Norbert
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I changed the question, and I think im within my rights too, in the original question I didn't just ask for a counter example. – Ethan Splaver Dec 27 '12 at 21:11
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@Norbert A quick test reveals the integral has to be between $0$ and $1$. $\lfloor 1/x \rfloor \leq 1/x$ and ${1/x} < 1$. Hence, we have $x \lfloor 1/x \rfloor {1/x} < x \times 1/x \times 1 = 1$. Hence, the integral has to be less than $1$. The divergence you have indicated would mean that the integral is negative, which is not possible. – Dec 27 '12 at 22:31
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3
Split the integral up into segments $S_m=[1/m,1/(m+1)]$ with $[0,1]= \cup_{m=1}^\infty S_m$. In the segment $m$, we have $\lceil 1/x \rceil=m+1$ and $\{1/x\} = 1/x- \lfloor 1/x\rfloor = 1/x - m$ (apart from values of $x$ on the boundary which do not contribute to the integral).
This yields $$\begin{align}\int_0^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx &= \sum_{m=1}^\infty \int_{S_m}x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx \\ &= \sum_{m=1}^\infty \int_{1/(m+1)}^{1/m} x (m+1)\left(\frac1x -m \right)\, dx\\ &= \sum_{m=1}^\infty \frac{1}{2m(1+m)}\\ &=\frac{1}{2}. \end{align}$$
Fabian
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