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Prove an identity for the sum of integrals of particular step functions

by RoRi

Prove

    \[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx = a-b. \]

Proof. From this exercise (1.11 #4, part b) we know

    \[ [-x] = \begin{cases} -[x] & \text{if } x \in \mathbb{Z} \\ -[x]-1 & \text{if } x \notin \mathbb{Z}. \end{cases} \]

But, since [x] is constant on the open subintervals of the partition

    \[ P = \{ a, [a] + 1, \ldots, [a] + [b-a], b \} \]

which contains every integer between a and b, we have [-x] = -[x]-1 on the open subintervals of P (since there are no integers in the open subintervals). Hence, \int_a^b [-x] \, dx = \int_a^b (-[x] -1)\, dx. Thus,

    \begin{align*} \int_a^b [x] \, dx + \int_a^b [-x] \, dx &= \int_a^b [x] \, dx + \int_a^b -[x]-1 \, dx \\ &= \int_a^b -1 \, dx \\ &= \int_b^a 1 \, dx & (\text{let } P = \{ a,b \})\\ &= a-b. \qquad \blacksquare \end{align*}

3 comments

  1. Prashant says:

    open subintervals of P does not contain integers, I agree. But why did you ignore integer points in [a,b] and concluded that [-x] = -[x]-1 always ?

  2. Trevor Garrity says:

    The partition does not contain all integers. For example, let a = 1.9 and b =3.8. [a] = 1, b-a = 1.9, and [b-a] = 1. Therefore, your proposed partition is 1.9, 2, 3.8. This is missing the integer 3.

    • Anonymous says:

      The partition does contain all integers including the integers in the edges of the interval, its the open subintervals of the Partition which define the subdivision points of the integration procedure, those are inmaterial.

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