Integration of Piecewise Functions

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

\[\int _{ 0 }^{ n }{ \left\lfloor x \right\rfloor \, dx }, \]

we can look at the following graph to visualize the integral:

The area under the function from \(0\) to \(x\) is \(0+1+2+3+\cdots+(x-1)\), and we can represent this as a sum

\[\int _{ 0 }^{ n }{ \left\lfloor x \right\rfloor \, dx } =\displaystyle\sum _{ k=0 }^{ n-1 }{ k } =\frac { n(n-1) }{ 2 }.\]

in general,\(\displaystyle\int _{ 0 }^{ nt }{ f(x)\, dx } =n\int _{ 0 }^{ t }{ f(x)\, dx }, \) where \(t\) is the period of the function.

Floor and ceiling functions are very basic and have simple patterns that their graphs follow; you can think of their graphs as stairs.

Find \(\displaystyle\int _{ 0 }^{ n }{ 7\left\lceil x \right\rceil \, dx }.\)

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This integral is equal to \(7\displaystyle\int _{ 0 }^{ n }{ \left\lceil x \right\rceil dx }.\)

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 \(0<x\le 1,\) then \(\left\lceil x \right\rceil =1\). Graphically that's a square with side 1.
If \(1<x\le 2,\) then \(\left\lceil x \right\rceil =2\), which is a rectangle with length 2 and width 1.
We can keep going on to verify the pattern.

Therefore, this integral is \(7\displaystyle\sum _{ k=1 }^{ n }{ k },\) which we can simplify it to get \(\boxed{\frac { 7n(n+1) }{ 2 }}.\) \(_\square\)

The graph of the given function is as follows:

Find \( \displaystyle\int _{ 1 }^{ \infty }{\frac{dx}{\left\lfloor x \right\rfloor} }. \)

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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 integral diverges. \(_\square\)

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 \(\displaystyle\int _{ 0 }^{ 6 }{ \left\{ x \right\}\, dx }.\)

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This integral can be written as \(\displaystyle\int _{ 0 }^{ 6 }{ \left\{ x \right\}\, dx } =\displaystyle\int _{ 0 }^{ 6 }{ \left(x-\left\lfloor x \right\rfloor \right) \, dx } =\displaystyle\int _{ 0 }^{ 6 }{ x }\, dx -\displaystyle\int _{ 0 }^{ 6 }{ \left\lfloor x \right\rfloor }\, dx. \)

We know how to calculate these integrals, which will make the answer 3. \(_\square\)

Graph of the funcion:

Evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} \, dx }.\)

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First we make the substitution \(t=\frac{1}{x}.\)

We then have the integral above 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. \(_\square\)

Prove that \(\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x+y }{ x-y } \right\}\, dx\, dy } } =\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x-y }{ x+y } \right\} \, dx\, dy } } =\frac { 1 }{ 2 }. \)

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By symmetry we have \(I=\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x-y }{ x+y } \right\} \, dx\, dy } } =\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { y-x }{ x+y } \right\} \, dx\, dy } }. \)

Thus \(I=\frac { 1 }{ 2 } (I+I)=\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \, dx\, dy } } =\frac { 1 }{ 2 }.\ _\square \)

Evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ x\left\lfloor \frac { 1 }{ x } \right\rfloor \left\{ \frac { 1 }{ x } \right\} \, dx. }\)

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Let \( \frac { 1 }{ x } =t,\) and the integral converts to

\[\begin{align} \int _{ 1 }^{ \infty }{ \frac { \{ t\} \left\lfloor t \right\rfloor }{ { t }^{ 3 } } dt } &=\sum _{ n=1 }^{ \infty }{ \displaystyle\int _{ n }^{ n+1 }{ \frac { (t-n)n }{ { t }^{ 3 } } dt } } \\ &=\frac { 1 }{ 2 } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ 2 } } } \\ &=\frac { 1 }{ 2 } \big( \zeta (2)-1 \big).\ _\square \end{align} \]

\[ \large \displaystyle\int _{ 0 }^{ 1 }{ \left \{ \dfrac { 1 }{ x } \right\} ^{ n } } \, dx=\sum _{ k=1 }^{ \infty }{ \frac { \zeta (k+1)-1 }{ \binom{n+k}{k} } } \]

Prove the equation above for positive integers \(n\).

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This is actually a result from a more generalized form \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } \big(\zeta (n+i+1)-1\big). } \)
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 \[\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ i }^{ i+1 }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } }\, dt } =\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (t-i)^{ k } }{ { t }^{ n+2 } } }. \] \(U\)-substituting by letting \(u=t-i,\) we get \[\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \frac { { u }^{ k } }{ (i+u)^{ n+2 } } } }\, du=\int _{ 0 }^{ 1 }{ { u }^{ k }\left( \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } \right) dy }. \qquad (1)\] Then since \( \frac { 1 }{ (i+u)^{ n+2 } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { e }^{ -(i+u)u }{ y }^{ n+1 } } dy, \) putting this in gives \[\begin{align} \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } &=\frac { 1 }{ (n+1)! } \sum _{ i=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -(i+u)y }{ y }^{ n+1 }\, dy } } \\ &=\frac { 1 }{ (n+1)! } \int _{ 0 }^{ \infty }{ { y }^{ n+1 } } { e }^{ -uy }\left(\sum _{ i=1 }^{ \infty }{ { e }^{ -iy } } \right)\, dy\\ &=\frac { 1 }{ (n+1)! } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy }.\qquad (2) \end{align}\] 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 \[\begin{align} \frac { 1 }{ (n+1)! } \sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+i }{ e }^{ -y } }{ { e }^{ y }-1 } \, dy } &= \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -2y }\sum _{ j=0 }^{ \infty }{ { e }^{ -jy } } dy } \\ &=\sum _{ j=0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -(2+j)y }\, dy }} \\ &=\sum _{ j=0 }^{ \infty }{ \frac { \Gamma (n+i+1) }{ (2+j)^{ n+i+1 } } } =(n+i)! \big( \zeta (n+i+1)-1 \big). \end{align}\] Hence, \[\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } (\zeta (n+i+1)-1) },\] and the result follows. \(_\square\)

\[ \int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { x }{ y } \right\} ^{ k }\dfrac { { y }^{ a } }{ { x }^{ b } } \, dx \; dy } } =\dfrac { 1 }{ a-b+2 } \left( \dfrac { 1 }{ k-b+1 } +\dfrac { k! }{ (a+1)! } \sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \Big( \zeta (a+n+1)-1 \Big) } \right) \]

Prove the equation above with \(k\) being a real number \(\ge1\) and \(a,b\) nonnegative integers such that \(a-b>-2\) and \(k-b>-1\).

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Rearranging the integral we get

\[\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ b } } \int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }{ y }^{ a }\, dx\, dy } }. \]

Then we let \(\frac { x }{ y } =u\) and the integral converts into

\[\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx.\]

Now we integrate by parts to get

\[\begin{align} \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx &=\left. \left( \frac { { x }^{ a+2-b } }{ a+2-b } \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right)\right|_{x=0}^{x=1} +\frac { 1 }{ a+2-b } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ k-b }dx } \\ &=\frac { 1 }{ a+2-b } \displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } +\frac { 1 }{ (a+2-b)(k-b+1) }. \end{align} \]

We evaluated that integral in the previous example.

Putting that in, we get

\[\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }\frac { { y }^{ a } }{ x^{ b } } \, dx\, dy } } =\frac { 1 }{ a-b+2 } \left( \frac { 1 }{ k-b+1 } +\frac { k! }{ (a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \Big( \zeta (a+n+1)-1 \Big) } \right).\ _\square\]

Evaluate \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { n }{ \sqrt [ n ]{ x } } \right\}\, dx }\) for \(n\ge2.\)

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We make the substitution \( x=\dfrac { { n }^{ n } }{ { y }^{ n } }, \) which will convert the integral to

\[\begin{align} { n }^{ n+1 }\displaystyle\int _{ n }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } dy } &={ n }^{ n+1 }\displaystyle\sum _{ k=n }^{ \infty }{ \displaystyle\int _{ k }^{ k+1 }{ \frac { y-k }{ { y }^{ n+1 } } dy } } \\ &=\frac { { n }^{ n+1 } }{ n-1 } \displaystyle\sum _{ k=n }^{ \infty }{ \left( \frac { 1 }{ { k }^{ n-1 } } -\frac { 1 }{ (k+1)^{ n-1 } } \right) } +{ n }^{ n }\displaystyle\sum _{ k=n }^{ \infty }{ k\left( \frac { 1 }{ (k+1)^{ n } } -\frac { 1 }{ { k }^{ n } } \right) } \\ &=\frac { { n }^{ 2 } }{ n-1 } +{ n }^{ n} \displaystyle\sum _{ k=n }^{ \infty }{ \left( \frac { 1 }{ (k+1)^{ n-1 } } -\frac { 1 }{ { k }^{ n-1 } } -\frac { 1 }{ (k+1)^{ n } } \right) } \\ &=\frac { n }{ n-1 } -{ n }^{ n }\left( \zeta (n)-\displaystyle\sum _{ k=1 }^{ n }{ \frac { 1 }{ { k }^{ n } } } \right).\ _\square \end{align}\]

Prove that \(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ n\sqrt [ n ]{ x } } \right\} dx } =\frac { 1 }{ n-1 } -\frac { \zeta (n) }{ { n }^{ n } } .\)

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First we let \(x=\frac { 1 }{ { n }^{ n }{ y }^{ n } },\) then the integral converts to

\[\frac { 1 }{ { n }^{ n-1 } } \displaystyle\int _{ \frac1n }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } } =\frac { 1 }{ { n }^{ n-1 } } \left( \displaystyle\int _{ \frac1n }^{ 1 }{ \frac { dy }{ { y }^{ n } } } +\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ y\} }{ { y }^{ n+1 } } dy } \right). \]

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

\[\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ n\sqrt [ n ]{ x } } \right\} dx } =\frac { 1 }{ n-1 } -\frac { \zeta (n) }{ { n }^{ n } }.\ _\square \]

add a solution if you find one

\[ \large \int_0^1 \int_0^1 \cdots \int_0^1 \left \{\frac{1}{\prod_{n=1}^k x_n} \right \} \, dx_1 \; dx_2 \; \cdots dx_k = 1 - \sum_{n=0}^{k-1} \dfrac{\gamma_n}{n!} \]

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.

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needs solution

\[ \large \int_0^1 \int_0^1 \cdots \int_0^1 \left\{\frac{1}{\sum_{n=1}^k x_n}\right\} ^m \, dx_1 \; dx_2 \cdots dx_k \]

The above is an unsolved problem in mathematics.
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.