6

So I was bored and decided to figure out the indefinite integral of the absolute value function, $|x|$. Using integration by parts ($u=|x|, dv=dx$, $dx = \text{sgn}(x)=\frac{|x|}{x}$), it can be shown that $\displaystyle\int |x| dx = \frac{x |x|}{2}+C$.

Now I decided to take the integral again, finding that $\displaystyle\int\left(\int |x| dx \right) dx=\frac{x^2 |x|}{3}+C$. Continuing, I found the pattern in the title, that the $n$th indefinite integral of $|x|$ is $\displaystyle\frac{x^n |x|}{n+1}+C$. Is there a way to prove this general result?

J. M. ain't a mathematician
  • 75,051
Eugene Bulkin
  • 1,827

3 Answers3

10

Yes, there is a way: Use Mathematical induction. I don't add more details because I think it's elementary, isn't it?

Rasmus
  • 18,404
  • 1
    Another easy way would be to (1) compute the derivative of $|x|$, then (2) compute the derivative of $|x|^n$ using the chain rule. – Ryan Budney Oct 09 '10 at 19:50
  • 1
    not sure why I didn't think of that... I'll try it out and see. – Eugene Bulkin Oct 09 '10 at 20:20
  • 1
    @Ryan: (1) you can't compute derivative of |x| at 0, (2) I don't see how |x|^n relates to this problem. – Marek Nov 11 '10 at 20:16
  • @Marek, it's not clear to me what you're referring to by "can't compute derivative of..." Perhaps you meant to say the derivative does not exist at $0$? This is not relevant to my point. My point is the fundamental theorem of calculus allows you to compute integrals via anti-derivatives. So by computing a derivative you in principle must be computing anti-derivatives. – Ryan Budney Nov 13 '10 at 18:00
  • @Ryan: well, if it does not exist then you surely can't compute it, so I think that's pretty relevant to what you said. Also, you didn't address the second point: why |x|^n? – Marek Nov 13 '10 at 18:13
  • @Marek: Compute its derivative and you should see precisely why I made the suggestion. FYI you seem to be mistaking "computable" for "derivative exists". These are different notions. – Ryan Budney Nov 13 '10 at 22:59
  • @Ryan: this is getting weird and we are clearly not able to communicate with each other (or at least I don't follow you at all). So I propose we let it be. – Marek Nov 13 '10 at 23:13
7

Make the observation that $|x| = \theta(x) x - \theta(-x) x$ for all real $x$, where $\theta$ is the Heaviside function (evaluating to $1$ if the argument is positive and $0$ otherwise). It is known that \begin{eqnarray} \int \theta(x) \ x \ dx = \theta(x) \frac{x^{2}}{2} + C \quad \text{and} \quad \int \theta(-x) \ x \ dx = \theta(-x) \frac{x^{2}}{2} + C^{\prime}, \end{eqnarray} where $C$ and $C^{\prime}$ are constants of integration. The identity for $n = 1$ follows by subtraction and the representation of $|x|$ above. With $n$ integrations, we have \begin{eqnarray} \int \cdots \int |x| \ dx = \theta(x) \int \cdots \int x \ dx - \theta(-x) \int \cdots \int x \ dx = \frac{|x| x^{n}}{(n+1)!} + P_{n}, \end{eqnarray} where $P_{n}$ is a polynomial in $x$, as claimed.

user02138
  • 17,064
  • I seen this message Heaviside function display to me before in my calculator. I wanted to ask what type of course(s) do they introduce this concept to students in because I have never heard about it yet in class. – night owl Jun 23 '11 at 09:57
  • @Night Owl: The Heaviside function $\theta(x)$ is defined in the following way: $\theta(x)$ is $0$ if $x < 0$ and $1$ otherwise. It is most often taught in a course of Differential Equations or Fourier Analysis. – user02138 Jun 23 '11 at 16:36
3

The $C$ in the title of this question should be replaced by an arbitrary polynomial $p(x)$ of degree $\le n-1$. After all, we are talking of an $n$-fold integration here.

Christian Blatter
  • 226,825