Integration of fractional part of x

Integration of fractional part of x

• The integration of the fractional part function, often denoted as {x} or frac(x), is a bit more complex than standard functions because the fractional part function is not continuous everywhere. The fractional part of x is defined as x - [x], where [x] is the floor function, representing the greatest integer less than or equal to x.
• The fractional part function has a sawtooth wave pattern and is periodic with a period of 1. To integrate this function over an interval, you typically break the interval into subintervals where the function is continuous.

For example, to integrate the fractional part function from a to b, where a and b are real numbers and a < b, you would do the following:

1. Identify the Intervals: Break the interval [a, b] into subintervals at each integer point. For example, if a = 1.5 and b = 4.5, the subintervals would be [1.5, 2], [2, 3], [3, 4], and [4, 4.5].
2. Integrate Over Each Subinterval: The integral of the fractional part function over each subinterval [n, n+1) (where n is an integer) is straightforward, as the function is simply x - n over this interval. So, the integral over [n, n+1) is (1/2)(n+1)^2 - (1/2)n^2.
3. Sum the Integrals: Add up the integrals over each subinterval.

For a general formula, if a and b are not integers, you would need to calculate the integrals over the first partial interval [a, ?a?] and the last partial interval [?b?, b] separately, and then sum the integrals over the full intervals in between.

Example 1
$\int_{1.5}^{4.5} \left\{ x \right \} dx$

Let’s integrate the fractional part of x from 1.5 to 4.5. We break this into intervals:

From 1.5 to 2:

• The fractional part function is x - 1.
• Integral: $\int_{1.5}^{2} (x – 1) dx$.

From 2 to 3:

• The fractional part function is x - 2.
• Integral: $\int_{2}^{3} (x – 2) dx$.

From 3 to 4:

• The fractional part function is x - 3.
• Integral: $\int_{3}^{4} (x – 3) dx$.

From 4 to 4.5:

• The fractional part function is x - 4.
• Integral: $\int_{4}^{4.5} (x – 4) dx$.

Now, calculate each integral:

1. $\int_{1.5}^{2} (x – 1) dx = \left[\frac{1}{2}x^2 – x\right]_{1.5}^{2} = \left(\frac{1}{2}(2)^2 – 2\right) – \left(\frac{1}{2}(1.5)^2 – 1.5\right) = .375$.
2. $\int_{2}^{3} (x – 2) dx = \left[\frac{1}{2}x^2 – 2x\right]_{2}^{3} = \left(\frac{1}{2}(3)^2 – 2(3)\right) – \left(\frac{1}{2}(2)^2 – 2(2)\right) =.5$.
3. $\int_{3}^{4} (x – 3) dx = \left[\frac{1}{2}x^2 – 3x\right]_{3}^{4} = \left(\frac{1}{2}(4)^2 – 3(4)\right) – \left(\frac{1}{2}(3)^2 – 3(3)\right)= .5$.
4. $\int_{4}^{4.5} (x – 4) dx = \left[\frac{1}{2}x^2 – 4x\right]_{4}^{4.5} = \left(\frac{1}{2}(4.5)^2 – 4(4.5)\right) – \left(\frac{1}{2}(4)^2 – 4(4)\right) =.125 $.

Finally, Sum is 1.5 units

Integration of the fractional part function using Periodic method

• The fractional part function is periodic with a period of 1. This means that the function repeats its values every interval of length 1. Mathematically, this is expressed as {x + n} = {x} for any integer n.
• When integrating the fractional part function over a full period (or multiple periods), the integral can be simplified due to its periodic nature. For example, the integral of {x} from 0 to 1 (a full period) can be calculated directly, and this result can be used to find integrals over longer intervals that span multiple periods.

$\int_{0}^{n} \left\{ x \right \} \; dx =n \int_{0}^{1} \left\{ x \right \} \; dx$

Now Since {x} equals x in this interval (as there are no integers between 0 and 1), the integral is straightforward

$\int_{0}^{n} \left\{ x \right \} \; dx =n \int_{0}^{1} \left\{ x \right \} \; dx = n \int_{0}^{1} x \; dx = n\left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{n}{2} $

Example
Evaluate

$ \int_{0}^{100} \left\{ x \right \} \, dx $
Solution
$$ \int_{0}^{100} \left\{ x \right \} \, dx= 100 \times \int_{0}^{1} \left\{ x \right \} \, dx = 100 \times \frac{1}{2} = \frac{100}{2}= 50 $$

Solved Examples

Question 1

$$ \int_{0}^{1} \left\{ x \right \} [x+1] \, dx $$
here {x} denotes fractional part and [.] denotes greatest integer function

Solution

$ \int_{0}^{1} \left\{ x \right\} [x+1] \, dx = \int_{0}^{1} (x – [x]) [x+1] \, dx$

now for $x \in (0,1)$, we have [x]=0 and [x+1] =1

Hence

$\int_{0}^{1} (x – [x]) [x+1] \, dx = \int_{0}^{1} x \, dx = \frac {1}{2} $