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I learned about integrating GIF functions if there is definite integrals of this form: $$\int_c ^d [x+1] dx$$ But need a bit help to integrate this GIF function if there is indefinite integral of: $$\int [x+1] dx$$

p.s I searched many places including Is it possible to integrate a greatest integer function? in different form but it's hard to understand.

Need a bit help on this Calc II problem if possible.

Thank you in advance :)

Zenit
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    Try to graph the function first: this should help significantly. – Jacob A Mar 28 '21 at 22:02
  • @John Tell us what you don't understand exactly in Rob Johns answer. – vitamin d Mar 28 '21 at 22:07
  • @JacobAdamczyk i draw graph and understand it but don't know how to evaluate the problem cause there are no borders, it's indefinite integral – Zenit Mar 28 '21 at 22:08
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    Do it for a few intervals [0,1],[1,2],[-1,0], then try a conjecture based on the pattern. As a hint your solution is an infinite sum. – Muselive Mar 28 '21 at 22:10
  • @Muselive thank you i did it for some intervals, how can I represent sum of all those infinite terms – Zenit Mar 28 '21 at 22:27
  • @John you don't have to. Just figure out the area under the curve. They're all rectangles so this should be nearly trivial. – CyclotomicField Mar 28 '21 at 22:36
  • @CyclotomicField all are giving different aproaches. I identified some areas of the rectangles too,there are infinite areas what should i do now. Which is considered to be the final answer of this kind of problem :( – Zenit Mar 28 '21 at 22:55
  • @john the bounds are $[a,b]$. You'll need to express the areas in terms of those two numbers. Feel free to use the floor and ceiling functions to express your answer. – CyclotomicField Mar 28 '21 at 23:14

1 Answers1

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Using Bernoulli Polynomials

Since the integral of $\{x\}-\tfrac12$ over a unit interval is $0$, we can write $$ \int\left(\{x\}-\tfrac12\right)\mathrm{d}x=\tfrac12\{x\}^2-\tfrac12\{x\}+C\tag1 $$ This is the beginning of the Bernoulli Polynomials. In fact, $B_1(x)=x-\frac12$ and $\frac12B_2(x)=\frac12x^2-\frac12x+\frac1{12}$.

Since $\lfloor x+1\rfloor=x+1-\{x+1\}=x+1-\{x\}$, we can use $(1)$ to get $$ \begin{align} \int\lfloor x+1\rfloor\,\mathrm{d}x &=\int\left(x+\tfrac12-\left(\{x\}-\tfrac12\right)\right)\,\mathrm{d}x\tag{2a}\\ &=\tfrac12x^2+\tfrac12x-\tfrac12\{x\}^2+\tfrac12\{x\}+C\tag{2b}\\[3pt] &=\tfrac12x^2+\tfrac12x-\tfrac12\left(x-\lfloor x\rfloor\right)^2+\tfrac12\left(x-\lfloor x\rfloor\right)+C\tag{2c}\\[3pt] &=x\lfloor x\rfloor-\tfrac12\lfloor x\rfloor^2+x-\tfrac12\lfloor x\rfloor+C\tag{2d} \end{align} $$

Using a Previous Answer

Since an answer to the referenced question was $$ \int\lfloor x\rfloor\,\mathrm{d}x=\lfloor x\rfloor x-\tfrac12\lfloor x\rfloor^2-\tfrac12[x]+C\tag3 $$ we can simply substitute $x\mapsto x+1$ to get $$ \begin{align} \int\lfloor x+1\rfloor\,\mathrm{d}x &=\color{#C00}{\lfloor x+1\rfloor(x+1)}\color{#090}{-\tfrac12\lfloor x+1\rfloor^2}\color{#00F}{-\tfrac12\lfloor x+1\rfloor}+C\tag{4a}\\ &=\color{#C00}{\lfloor x\rfloor x+x+\lfloor x\rfloor+1}\color{#090}{-\tfrac12\lfloor x\rfloor^2-\lfloor x\rfloor-\tfrac12}\color{#00F}{-\tfrac12\lfloor x\rfloor-\tfrac12}+C\tag{4b}\\[3pt] &=\lfloor x\rfloor x-\tfrac12\lfloor x\rfloor^2+x-\tfrac12\lfloor x\rfloor+C\tag{4c} \end{align} $$

robjohn
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  • Thank you so much, wish you all the best :) – Zenit Apr 03 '21 at 14:09
  • Completely understandable elaboration, but can't get this one $x+1-{x+1}=x+1-{x}$ sorry for my late question :) – Zenit Apr 03 '21 at 20:26
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    Don't worry about late questions. Some comments are posted years after the original post. Since $1\in\mathbb{Z}$, the fractional part of $x$ is the same as the fractional part of $x+1$. Think of the fractional part of $3.14159265358979$ and the fractional part of $4.14159265358979$ – robjohn Apr 03 '21 at 21:04