Integration of Piecewise Functions
Indefinite integrals of floor,celing.and fractional have a closed form,but this condition might not hold at sometimes,and it's way easier to not try to find the definite integral and directly proceeding 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
\(\displaystyle\int _{ 0 }^{ n }{ \left\lfloor x \right\rfloor \quad dx } \)
we can look at the graph to visualize the integral
the area under the function from \(0\) to \(x\) is \(0+1+2+3..+(x-1)\) ,we can represent this as a sum
\(\displaystyle\int _{ 0 }^{ n }{ \left\lfloor x \right\rfloor \quad dx } =\displaystyle\sum _{ k=0 }^{ n-1 }{ k } =\frac { n(n-1) }{ 2 }\)
in general,\(\displaystyle\int _{ 0 }^{ nt }{ f(x)dx } =n\int _{ 0 }^{ t }{ f(x)dx } \)
where \(t\) is the period of the function
Contents
Floor and Ceiling Functions
Floor and ceiling functions are very basic,and have simple patterns that their graphs follow,you can think of their graphs as stairs
find \(\displaystyle\int _{ 0 }^{ n }{ 7\left\lceil x \right\rceil dx } \\ \)
this integral is equal to \(7\displaystyle\int _{ 0 }^{ n }{ \left\lceil x \right\rceil dx } \\ \)
now we need to think about the sum generated, we can graph the function,and it'll be easier,but let us think about it without graphing
if \(0<x\le 1 \) then \(\left\lceil x \right\rceil =1\),graphically that's a square with side 1 if \(1<x\le 2\) then \(\left\lceil x \right\rceil =2\),this is a rectangle with length \(2\) and width \(1\)
we can keep going on to verify the pattern
this integral is \(7\displaystyle\sum _{ k=1 }^{ n }{ k } \) simplifying we get the integral is \(\boxed{\frac { 7n(n+1) }{ 2 }} \)
graph of the function
\[ \large \int_0^4 \lceil x-1 \rceil \, dx = \, ? \]
Notation: \( \lceil \cdot \rceil \) denotes the ceiling function.
find \( \displaystyle\int _{ 1 }^{ \infty }{\frac{1}{\left\lfloor x \right\rfloor} } \)
if we try to represent this as a sum,it'll be \(\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } } \)
we know this sum diverges,so the answer is that the \(\boxed{\text{integral diverges}}\)
\[ \large \displaystyle\int _{ 1 }^{ \infty }{ \dfrac { dx }{ \lfloor x \rfloor ^{ 2 } } } =\dfrac { { \pi }^{ a } }{ b } \]
The equation above holds true for positive integers \(a\) and \(b\), find \(a+b\).
Notation: \( \lfloor \cdot \rfloor \) denotes the floor function.
\[ \large \displaystyle\int _{ 0 }^{ 2 }{ \left\lfloor { x }^{ 2 }-x \right\rfloor \, dx } =a-b\phi \]
The equation above holds true for positive integers \(a\) and \(b\), and \(\phi\) denotes the Golden ratio, \(\phi = \dfrac{1+\sqrt5}2 \). Find \(a+b\).
Fractional Part Integrals
the fractional part function,denoted by \(\left\{ x \right\} \) is the decimal part of the number \(x\) or in mathematical terms,\(\left\{ x \right\} =x-\left\lfloor x \right\rfloor \)
find \(\displaystyle\int _{ 0 }^{ 6 }{ \left\{ x \right\} dx } \)
this integral can be written as \(\displaystyle\int _{ 0 }^{ 6 }{ \left\{ x \right\} dx } =\displaystyle\int _{ 0 }^{ 6 }{ x-\left\lfloor x \right\rfloor dx } =\displaystyle\int _{ 0 }^{ 6 }{ x } -\displaystyle\int _{ 0 }^{ 6 }{ \left\lfloor x \right\rfloor } \)
we know how to calculate these integrals,which will make the answer \(\boxed{3}\) graph of the funcion:
\[ \large \int_0^4 \{ x \} \lfloor x \rfloor \, dx = \, ? \]
Notations:
\( \{ \cdot \} \) denotes the fractional part function.
\( \lfloor \cdot \rfloor \) denotes the floor function.
Intermediate Examples
evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} dx } \)
first we make the substitution \(t=\frac{1}{x}\)
we then have the integral above is equal to
\(\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} }{ { t }^{ 2 } } dt } \)
we then convert it into a sum of consecutive integrals
\(\displaystyle\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { t-n }{ { t }^{ 2 } } dt } } \) this integral is easy to evaluate,and will result in the summation \(\displaystyle\sum _{ n=1 }^{ \infty }{ \left( \ln { \left( \frac { n+1 }{ n } \right) } -\frac { 1 }{ n+1 } \right) } =1-\gamma \) where \(\gamma\) is the Euler-Mascheroni constant
this is a very famous and beautiful result in mathematics
\[\large \int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ { x }^{1/6 } } \right\} \,dx } =\dfrac { A }{ B } -\dfrac { { \pi }^{ C } }{ D } \]
The equation above holds true for positive integers \(A,B,C\) and \(D\), with \(A,B\) coprime. Find \(A+B+C+D\).
Notation: \( \{ \cdot \} \) denotes the fractional part function.
\[ \large \int_1^\infty \dfrac{\{ x \}} {x^5} \, \mathrm{d}x \]
If the value of the integral above is equal to \(\dfrac1A - \dfrac{\pi^B}C \) for positive integers \(A\), \(B\) and \(C\), find \(A+B+C\).
Bonus: Find the general form of \( \displaystyle \int_1^\infty \dfrac{\{x\}}{x^n} \, \mathrm{d}x \).
Clarification: \(\{x\}\) denotes the fractional part function.
prove that \(\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x+y }{ x-y } \right\} dxdy } } =\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x-y }{ x+y } \right\} dxdy } } =\frac { 1 }{ 2 } \)
by symmetry we have \(I=\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x-y }{ x+y } \right\} dxdy } } =\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { y-x }{ x+y } \right\} dxdy } } \)
thus \(I=\frac { 1 }{ 2 } (I+I)=\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ dxdy } } =\frac { 1 }{ 2 } \)
evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ x\left\lfloor \frac { 1 }{ x } \right\rfloor \left\{ \frac { 1 }{ x } \right\} dx }\)
let \( \frac { 1 }{ x } =t\) and the integral converts to
\(\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} \left\lfloor t \right\rfloor }{ { t }^{ 3 } } dt } =\displaystyle\sum _{ n=1 }^{ \infty }{ \displaystyle\int _{ n }^{ n+1 }{ \frac { (t-n)n }{ { t }^{ 3 } } dt } } =\frac { 1 }{ 2 } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ 2 } } } =\\ \frac { 1 }{ 2 } \left( \zeta (2)-1 \right) \)
Advanced Problems
\[ \Large \displaystyle\int _{ 0 }^{ 1 }{ \left \{ \dfrac { 1 }{ x } \right\} ^{ n } } \, dx=\sum _{ k=1 }^{ \infty }{ \frac { \zeta (k+1)-1 }{ \binom{n+k}{k} } } \] Prove the equation above for positive integer \(n\).
this is actually a result from a more generalized form, \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } (\zeta (n+i+1)-1) } \)
let's denote the above as
\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =f(k,n)=\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } dt } \)
converting this to a sum of integrals we have
\(\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ i }^{ i+1 }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } } dt } =\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (t-i)^{ k } }{ { t }^{ n+2 } } } \)
u sub : \(u=t-i\) we get \(\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \frac { { u }^{ k } }{ (i+u)^{ n+2 } } } } du=\int _{ 0 }^{ 1 }{ { u }^{ k }\left( \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } \right) dy } (1)\) then i tried coming with an integral to represent the sum,after some searching i got \( \frac { 1 }{ (i+u)^{ n+2 } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { e }^{ -(i+u)u }{ y }^{ n+1 } } dy \) putting that in we get
\(\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } =\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -(i+u)y }{ y }^{ n+1 }dy } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { y }^{ n+1 } } { e }^{ -uy }(\displaystyle\sum _{ i=1 }^{ \infty }{ { e }^{ -iy } } )dy=\\ \frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } (2)\)
combining (1) and (2) we get \(\frac { 1 }{ (n+1)! } \int _{ 0 }^{ 1 }{ { u }^{ k }\left( \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } \right) du } =\frac { 1 }{ (n+1)! } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 } }{ { e }^{ y }-1 } } \left( \int _{ 0 }^{ 1 }{ { u }^{ k }{ e }^{ -uy }du } \right) dy\\ \\ \)
by IBP we can evaluate the inner integral as \(k!{ e }^{ -y }\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { { y }^{ i-1 } }{ (k+i)! } } \) so this simplifies to
\( \frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+i }{ e }^{ -y } }{ { e }^{ y }-1 } dy } \)
evaluating that integral we have
\(\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+i }{ e }^{ -y } }{ { e }^{ y }-1 } dy }= \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -2y }\displaystyle\sum _{ j=0 }^{ \infty }{ { e }^{ -jy } } dy } =\sum _{ j=0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -(2+j)y }dy } =\\ \displaystyle\sum _{ j=0 }^{ \infty }{ \frac { \Gamma (n+i+1) }{ (2+j)^{ n+i+1 } } } } =(n+i)!(\zeta (n+i+1)-1)\)
hence,
\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } (\zeta (n+i+1)-1) }\)
and the result follows
QED
\[ \large\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { x }{ y } \right\} ^{ k }\dfrac { { y }^{ a } }{ { x }^{ b } } \, dx \; dy } } =\dfrac { 1 }{ a-b+2 } \left( \dfrac { 1 }{ k-b+1 } +\dfrac { k! }{ (a+1)! } \sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \left( \zeta (a+n+1)-1 \right) } \right) \] Prove the equation above with \(k\) being a real number \(\ge1\) and \(a,b\) nonnegative integers such that \(a-b>-2\) and \(k-b>-1\).
rearranging the integral we get
\(\displaystyle\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ b } } \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }{ y }^{ a }dxdy } } \)
then we let \(\frac { x }{ y } =u\) and the integral converts into
\(\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx\)
now we integrate by parts to get
\(\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx=\left( \frac { { x }^{ a+2-b } }{ a+2-b } \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right)\Big|_{x=0}^{x=1} +\frac { 1 }{ a+2-b } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ k-b }dx } =\\ \frac { 1 }{ a+2-b } \displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } +\frac { 1 }{ (a+2-b)(k-b+1) } \)
we evaluated that integral in the previous example
putting that in we get
\(\boxed{\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }\frac { { y }^{ a } }{ x^{ b } } dxdy } } =\frac { 1 }{ a-b+2 } \left( \frac { 1 }{ k-b+1 } +\frac { k! }{ (a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \left( \zeta (a+n+1)-1 \right) } \right)}\)
\[ \large \int_0^1 \left \{ \dfrac{(-1)^{\lfloor 1/x\rfloor }}{x}\right \} \, dx = A + B \ln \left( \dfrac C{\pi} \right) \]
The equation above holds true for positive integers \(A,B\) and \(C\). Find \(A+B+C\).
Notation: \( \{ \cdot \} \) denotes the fractional part function.
this is a part of Who's up to the challenge?
\[\large \int_0^1 \int _{ 0 }^{ 1 }{ { \left\{ \dfrac { x }{ y } \right\} }^{ 3 }{ \left\{ \dfrac { y }{ x } \right\} }^{ 3 } \, dx \; dy } =A-\dfrac { { \pi }^{ B } }{ C } -\dfrac { { \pi }^{ B+2 } }{ D } -\zeta \left( E \right) \]
The equation above holds true for positive integers \(A,B,C,D\) and \(E\). Find \(A+B+C+D+E\).
Notations:
\( \{ \cdot \} \) denotes the fractional part function.
\(\zeta(\cdot) \) denotes the Riemann zeta function.
\[ \large \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ x } \right\} \dfrac { x }{ 1-x } \, dx=A\gamma } +B\]
The equation above holds true for integers \(A\) and \(B\). Find \(A+B\).
Notations:
\( \{ \cdot \} \) denotes the fractional part function.
\( \gamma\) denote the Euler-Mascheroni constant, \(\gamma \approx 0.5772 \).
\[ \large \displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ x\left\{ \dfrac { 1 }{ 1-xy } \right\} \, dx\; dy } } =A-\dfrac { \zeta (B) }{ C } \]
If the equation above holds true for positive integers \(A,B\) and \(C\), find \(A\times B \times C\).
Notations:
\( \{ \cdot \} \) denotes the fractional part function.
\(\zeta(\cdot) \) denotes the Riemann zeta function.
This is a part of Who's up to the challenge?
\[ \large \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ { x }^{ 1729 } } -\dfrac { 1 }{ (1-x)^{ 1729 } } \right\} { x }^{ 6 }(1-x)^{ 6 } \, dx } =\dfrac { a}{ b } \]
The equation above holds true for coprime positive integers \(a\) and \(b\). Find \(a+b\).
Bonus: Generalize it.
Notation: \( \{ \cdot \} \) denotes the fractional part function.
This problem is a part of set: Who's up to the challenge?
\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ 3 } } dx=−\frac { H }{ U } −Mγ+\frac { { M }_{ 1 } }{ U_1 } \ln { (Sπ) } −B\ln { A } \)
here,\(A\) is the Glaisher–Kinkelin constant
all the other variables in here are positive integers and all the fractions mentioned are coprime
find \(H+U+M+{ M }_{ 1 }+{ U }_{ 1 }+S+B\)
Note:\(\{x\}\) denotes the fractional part of \(x\)
Congratulate yourself if you solved this!
this is a part of Who's up to the challenge?
evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { n }{ \sqrt [ n ]{ x } } \right\} dx } = \) for \(n\ge2\)
first we make the substitution \( x=\frac { { n }^{ n } }{ { y }^{ n } } \)
which will convert the integral to
\( { n }^{ n+1 }\displaystyle\int _{ n }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } dy } ={ n }^{ n+1 }\displaystyle\sum _{ k=n }^{ \infty }{ \displaystyle\int _{ k }^{ k+1 }{ \frac { y-k }{ { y }^{ n+1 } } dy } } =\frac { { n }^{ n+1 } }{ n-1 } \displaystyle\sum _{ k=n }^{ \infty }{ \left( \frac { 1 }{ { k }^{ n-1 } } -\frac { 1 }{ (k+1)^{ n-1 } } \right) } +{ n }^{ n }\displaystyle\sum _{ k=n }^{ \infty }{ k\left( \frac { 1 }{ (k+1)^{ n } } -\frac { 1 }{ { k }^{ n } } \right) } =\\ \frac { { n }^{ 2 } }{ n-1 } +{ n }^{ n} \displaystyle\sum _{ k=n }^{ \infty }{ \left( \frac { 1 }{ (k+1)^{ n-1 } } -\frac { 1 }{ { k }^{ n-1 } } -\frac { 1 }{ (k+1)^{ n } } \right) } \\ =\frac { n }{ n-1 } -{ n }^{ n }\left( \zeta (n)-\displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ { k }^{ n } } } \right) \)
prove that \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ n\sqrt [ n ]{ x } } \right\} dx } =\frac { 1 }{ n-1 } -\frac { \zeta (n) }{ { n }^{ n } } \)
first we let \(x=\frac { 1 }{ { n }^{ n }{ y }^{ n } }\)
then the integral converts to
\(\frac { 1 }{ { n }^{ n-1 } } \displaystyle\int _{ 1/n }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } } =\frac { 1 }{ { n }^{ n-1 } } \left( \displaystyle\int _{ 1/n }^{ 1 }{ \frac { dy }{ { y }^{ n } } } +\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } dy } \right) \)
the first integral is easy to evaluate and we evaluated the second here (see my solution-Hummus a)
so with a bit of simplification we obtain the result
\(\boxed{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ n\sqrt [ n ]{ x } } \right\} dx } =\frac { 1 }{ n-1 } -\frac { \zeta (n) }{ { n }^{ n } }} \)
\[\int_0^1 \sqrt[3]{\frac{\{1/x\}}{1-\{1/x\}}}\frac{dx}{1-x} \]
If the closed form of the value of the integral above can be expressed can be expressed as \(\dfrac{a\pi^k}{c\sqrt{d}}\) where \(a\) and \(c\) are coprime and \(d\) is square-free. Find \(a+k+c+d\).
Also, is it possible to find the following in a closed form?
\[\int_0^1 \sqrt[n]{\frac{\{1/x\}}{1-\{1/x\}}}\frac{dx}{1-x} \]
Very Advanced Problems
add a solution if you find one
\[ \large \int_0^1 \int_0^1 \cdots \int_0^1 \left \{1 \div \prod_{n=1}^k x_n \right \} \, dx_1 \; dx_2 \; \cdots dx_k = 1 - \sum_{n=0}^{k-1} \dfrac{\gamma_n}{n!} \]
Prove that the equation above holds true where \( \gamma_n \) denotes the \(n^\text{th} \) Stieltjes constant, \(\displaystyle \gamma_n := \lim_{m\to\infty} \left [ \sum_{k=1}^m \dfrac{(\ln k)^n}k - \dfrac{(\ln m)^{n+1}}{n+1}\right ] \).
Notation: \( \{ \cdot \} \) denotes the fractional part function.
Clarification: There are \(k\) integrals.
needs solution
this is an unsolved problem in mathematics
\[ \large \int_0^1 \int_0^1 \cdots \int_0^1 \left\{ 1\div \sum_{n=1}^k x_n \right\} ^m \, dx_1 \; dx_2 \cdots dx_k \] Calculate in closed form, the integral above for \(k\geq 3\) and \(m\geq1 \).
Notation: \( \{ \cdot \} \) denotes the fractional part function.
Clarification: There are \(k\) integrals.