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I am trying to calculate following integral:

$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$

I tried usual change t=1/x but not able to further advance.

Thanks!

hmakholm left over Monica
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24th_moonshine
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    Hint: try to write it as a sum, by breaking the interval into subintervals wherein the floor function is constant. – lulu Oct 01 '18 at 16:53
  • @lulu you mean ceiling not floor – user10354138 Oct 01 '18 at 17:03
  • @user10354138 True, thanks. – lulu Oct 01 '18 at 17:05
  • Using $\lceil y \rceil = y+1-{y}$ where ${.}$ is the fractional part brings this problem back to problems treated here recently, such as https://math.stackexchange.com/questions/2905488/closed-form-of-integral-over-fractional-part-int-01-left-frac12-leftx – Dr. Wolfgang Hintze Oct 01 '18 at 17:50

3 Answers3

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$$\int_{1}^{\infty}\frac{\lceil x\rceil}{x^3}\,dx =\sum_{n\geq 1}\int_{n}^{n+1}\frac{n+1}{x^3}\,dx=\sum_{n\geq 1}\frac{2n+1}{2n^2(n+1)}=\color{red}{\frac{1}{2}+\frac{\pi^2}{12}}$$ holds by partial fraction decomposition.

Jack D'Aurizio
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If $\frac 1{n} \le x < \frac 1{n-1}$ then $\lceil \frac 1x \rceil = n$. and $\int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx= \int_{\frac 1{n}}^{\frac 1{n-1}} xn dx = n\frac {x^2}2|_{\frac 1n}^{\frac 1{n-1}}= \frac n2(\frac 1{(n-1})^2 -\frac 1{n^2}=\frac n2(\frac {2n-1}{n^2(n-1)^2})=\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}$

So $\int_{0}^{1} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty \int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty[\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}]=\sum\limits_{n=1}^\infty[\frac 1{n^2} - \frac {1}{2(n+1)n^2}]$

fleablood
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$$ \begin{aligned} \int_0^1x\cdot\left\lceil \frac{1}{x}\right\rceil\; dx &= \sum_{k\ge 1} \int_{1/(k+1)}^{1/k}x\cdot \underbrace{\left\lceil \frac{1}{x}\right\rceil}_{\in (k,k+1)\text{ a.e}} \;dx \\ &= \sum_{k\ge 1} \int_{1/(k+1)}^{1/k}x\cdot (k+1)\;dx \\ &= \sum_{k\ge 1} (k+1) \left[\ \frac 12 x^2\ \right]_{1/(k+1)}^{1/k} \\ &= \frac 12 \sum_{k\ge 1} (k+1) \left[\ \frac 1{k^2}-\frac 1{(k+1)^2}\ \right]_{1/(k+1)}^{1/k} \\ &= \frac 12 \sum_{k\ge 1} \frac {2k+1}{k^2(k+1)} \\ &= \frac 12 \sum_{k\ge 1} \frac k{k^2(k+1)} + \frac 12 \sum_{k\ge 1} \frac {k+1}{k^2(k+1)} \\ &= \frac 12 \sum_{k\ge 1} \left[ \frac 1k-\frac 1{k+1}\right] + \frac 12 \sum_{k\ge 1} \frac 1{k^2} \\ &= \frac 12\cdot 1+\frac 12\cdot\frac{\pi^2}6\ . \end{aligned} $$ Computer check, here PARI/GP:

? intnum( x=1.e-12, 1, x*ceil(1/x), 8 )
%12 = 1.3224146303918600341384604269228003273
? 1/2 + Pi^2/12
%13 = 1.3224670334241132182362075833230125946
dan_fulea
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