Indefinite integrals of floor, ceiling, and fractional part functions each have a closed form, but this condition might not hold sometimes, and it's way easier to not try to find the definite integral but directly proceed to solve the indefinite integral. The way to think about integrating these types of functions is by thinking of them as a sum. For example, for the integral
∫0n⌊x⌋dx,
we can look at the following graph to visualize the integral:
The area under the function from 0 to x is 0+1+2+3+⋯+(x−1), and we can represent this as a sum
∫0n⌊x⌋dx=k=0∑n−1k=2n(n−1).
in general,∫0ntf(x)dx=n∫0tf(x)dx, where t is the period of the function.
Floor and ceiling functions are very basic and have simple patterns that their graphs follow; you can think of their graphs as stairs.
Find ∫0n7⌈x⌉dx.
This integral is equal to 7∫0n⌈x⌉dx.
We now need to think about the sum generated, and if we graph the function, it'll be easier. But let's think about it without graphing.
If 0<x≤1, then ⌈x⌉=1. Graphically that's a square with side 1.
If 1<x≤2, then ⌈x⌉=2, which is a rectangle with length 2 and width 1.
We can keep going on to verify the pattern.
Therefore, this integral is 7k=1∑nk, which we can simplify it to get 27n(n+1).□
The graph of the given function is as follows:
∫04⌈x−1⌉dx=?
Notation: ⌈⋅⌉ denotes the ceiling function.
Find ∫1∞⌊x⌋dx.
If we try to represent this as a sum, it'll be n=1∑∞n1.
We know this sum diverges, so the answer is that the integral diverges. □
∫1∞⌊x⌋2dx=bπa
The equation above holds true for positive integers a and b. Find a+b.
Notation: ⌊⋅⌋ denotes the floor function.
∫02⌊x2−x⌋dx=a−bϕ
The equation above holds true for positive integers a and b, where ϕ=21+5 is the golden ratio.
Find a+b.
Fractional Part Integrals
The fractional part function, denoted by {x}, is the decimal part of the number x, or in mathematical terms, {x}=x−⌊x⌋.
Find ∫06{x}dx.
This integral can be written as ∫06{x}dx=∫06(x−⌊x⌋)dx=∫06xdx−∫06⌊x⌋dx.
We know how to calculate these integrals, which will make the answer 3. □
Graph of the funcion:
∫04{x}⌊x⌋dx=?
Notations:
{⋅} denotes the fractional part function.
⌊⋅⌋ denotes the floor function.
Intermediate Examples
Evaluate ∫01{x1}dx.
First we make the substitution t=x1.
We then have the integral above equal to
∫1∞t2{t}dt.
We then convert it into a sum of consecutive integrals:
n=1∑∞∫nn+1t2t−ndt.
This integral is easy to evaluate, and will result in the summation n=1∑∞(ln(nn+1)−n+11)=1−γ, where γ is the Euler-Mascheroni constant.
This is a very famous and beautiful result in mathematics. □
∫01{x1/61}dx=BA−DπC
The equation above holds true for positive integers A,B,C, and D, with A and B coprime.
Find A+B+C+D.
Notation: {⋅} denotes the fractional part function.
∫1∞x5{x}dx
If the value of the integral above is equal to A1−CπB for positive integers A, B and C, find A+B+C.
Bonus: Find the general form of ∫1∞xn{x}dx.
Clarification: {x} denotes the fractional part function.
∫1∞x3{x}dx=a−cπb
The above integral is true for positive integers a,b, and c.
What is the value of a+b+c?
Note:{x} denotes the fractional part of x.
∫01{x1}2017dx
If the value of the above integral can be represented as
j=1∑∞(jC+j)ζ(j+A)−B
for positive integers A,B and C, then evaluate A+B+C.
Notations:
{⋅} denotes the fractional part function.
ζ(⋅) denotes the Riemann zeta function.
(NM)=N!(M−N)!M! denotes the binomial coefficient.
Hint: Generalize for
∫01{x1}kdx
where k≥0 is an integer.
For more problems on calculus, click here.
∫01{20172017x1}dx
If the closed form of the integral above can be represented as b1−deζ(c), where a and b are co-prime integers, then evaluate b+c+d+e.
Notations
{⋅} denotes the fractional part function.
ζ(⋅) denotes the Reimann Zeta Function.
Inspiration Tapas Mazumdar.
Try this first.
∫01{2017x1}dx
If the value of the above integral can be represented in the closed form as ba−ζ(c) where a and b are co-prime integers, then evaluate a+b+c.
Notations:
{⋅} denotes the fractional part function.
ζ(⋅) denotes the Riemann zeta function.
Check out this problem if you liked this one.
For more problems on calculus, click here.
Prove that ∫01∫01{x−yx+y}dxdy=∫01∫01{x+yx−y}dxdy=21.
By symmetry we have I=∫01∫01{x+yx−y}dxdy=∫01∫01{x+yy−x}dxdy.
This is actually a result from a more generalized form ∫01{x1}kxndx=(n+1)!k!i=1∑∞(k+i)!(n+i)!(ζ(n+i+1)−1).
Let's denote the above as
∫01{x1}kxndx=f(k,n)=∫1∞tn+2{t}kdt.
Converting this to a sum of integrals, we have
i=1∑∞∫ii+1tn+2{t}kdt=i=1∑∞tn+2(t−i)k.U-substituting by letting u=t−i, we get
i=1∑∞∫01(i+u)n+2ukdu=∫01uk(i=1∑∞(i+u)n+21)dy.(1)
Then since (i+u)n+21=(n+1)!1∫0∞e−(i+u)uyn+1dy, putting this in gives
i=1∑∞(i+u)n+21=(n+1)!1i=1∑∞∫0∞e−(i+u)yyn+1dy=(n+1)!1∫0∞yn+1e−uy(i=1∑∞e−iy)dy=(n+1)!1∫0∞ey−1yn+1e−uydy.(2)
Combining (1) and (2), we get
(n+1)!1∫01uk(∫0∞ey−1yn+1e−uydy)du=(n+1)!1∫0∞ey−1yn+1(∫01uke−uydu)dy.
By IBP we can evaluate the inner integral as k!e−yi=1∑∞(k+i)!yi−1, so this simplifies to
(n+1)!1i=1∑∞(k+i)!k!∫0∞ey−1yn+ie−ydy.
Evaluating that integral, we have
(n+1)!1i=1∑∞(k+i)!k!∫0∞ey−1yn+ie−ydy=∫0∞yn+ie−2yj=0∑∞e−jydy=j=0∑∞∫0∞yn+ie−(2+j)ydy=j=0∑∞(2+j)n+i+1Γ(n+i+1)=(n+i)!(ζ(n+i+1)−1).
Hence,
∫01{x1}kxndx=(n+1)!k!i=1∑∞(k+i)!(n+i)!(ζ(n+i+1)−1),
and the result follows. □