How would I compute this integral with a ceiling function?

Question

It seems simple, but it is not, really. How would I calculate this ($\lceil a \rceil$ denotes the ceiling function)? $$\int{\lceil x+2 \rceil}\ln x\,\,dx$$

First, I noticed that $\lceil x+2 \rceil=\lceil x \rceil + 2$. So I got: $$\int({\lceil x \rceil}+2)\ln x\,\,dx=\int2\ln x\,\,dx+\int{\lceil x \rceil}\ln x\,\,dx$$ Then I got: $$\int2\ln x\,\,dx=2\int\ln x\,\,dx=2x\ln x-2x$$ So then: $$2x\ln x-2x+\int{\lceil x \rceil}\ln x\,\,dx$$ From this point, I tried integrating by parts. So first, I substituted: $$u=\ln x\qquad dv=\lceil x \rceil$$$$du=\frac1x\qquad v=\int\lceil x \rceil=\,?$$

So I'm left with: $$2x\ln x-2x+\Bigg(\ln x\int\lceil x\rceil-\int\frac1x\Bigg(\int\lceil x\rceil\Bigg)\Bigg)$$ But now I'm stuck. Am I doing correct so far? How do I even calculate the integral of $\lceil x\rceil$?

Answer 1

$$ \begin{align} \int\lceil x+2\rceil\log(x)\,\mathrm{d}x &=\int\lceil x+2\rceil\,\mathrm{d}(x\log(x)-x)\\ &=\lceil x+2\rceil(x\log(x)-x)-\int(x\log(x)-x)\,\mathrm{d}\lceil x+2\rceil\\ &=\lceil x+2\rceil(x\log(x)-x)-\sum_{k=1}^{\lceil x-1\rceil}(k\log(k)-k)+C \end{align} $$

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$$ \begin{align} \int\lceil x\rceil\,\mathrm{d}x &=x\lceil x\rceil-\int x\,\mathrm{d}\lceil x\rceil\\ &=x\lceil x\rceil-\sum_{k=0}^{\lceil x-1\rceil}k+C\\ &=x\lceil x\rceil-\frac{\lceil x\rceil^2-\lceil x\rceil}2+C \end{align} $$

Answer 2

$$\int\lceil x\rceil\,dx=x\lceil x\rceil- T_{\lceil x\rceil-1}+C,$$ where $T_n $ is the $n $th triangle number. Between integers the integrand is constant, so we expect the integral to increase linearly. We expect the rate of increase to be equal to the locally constant value of the integrand. We have to subract the triangle numbers so that the integral is continuous.