# Evaluating an Indefinite Integral involving $\{x\} = x - \lfloor x \rfloor$

This is probably a pretty simple question, but I just want to check something that I'm not completely sure about. I want to evaluate $\int{\{x\}}^ndx$, with $\{x\} = x - \lfloor x \rfloor$. Instead of using substitution or anything like that, I instead simply looked for a function that measures the area under $\{x\}^n$. The function (as seen here for $n = 4$), is just slices of $y = x^n$ with $x \in[0, 1]$. So the area from $0$ to $x$ is given by

$$\lfloor x \rfloor\int{x^ndx} + \{x\}\int{x^ndx} = (\lfloor x \rfloor + \{x\})\int{x^ndx} = x\int{x^ndx} = \frac{x^{n+2}}{n+1} + C.$$

But since the area under $\{x\}^n$ is always positive, and $x^{n+2}$ is negative when $n+2$ is odd, the final antiderivative should actually be

$$\int{\{x\}^ndx} = \frac{|x^{n+2}|}{n+1} + C, n \neq -1$$

Is it wrong to do this? Should I be evaluating indefinite integrals solely using the tools created to evaluate them (such as substitution, integration by parts, etc.) or are geometric proofs for the value of indefinite integrals equally valid?

Also I have a strange feeling in my gut that this is really wrong for some reason, like instead of geometrically finding a function that measures the area under $\{x\}^n$ I should have been looking for a function whose slope is $\{x\}^n$, so can someone clarify whether the antiderivative I got is even correct or not?

Because $\{n+x\}^n=x^n$ where $x \in [0,1), n \in \mathbb Z$, this function is periodic and is defined by its values in $[0,1)$. So if $x \ge 0,\ n \ne -1$:

\begin{align} \int_0^x\{t\}^n\ dt&=\int_0^{\lfloor x \rfloor+\{x\}}\{t\}^n\ dt\\ &=\int_0^{\lfloor x \rfloor}\{t\}^n \ dt+\int_{\lfloor x \rfloor}^{\lfloor x \rfloor+\{x\}}\{t\}^n\ dt\\ &=\int_0^{\lfloor x \rfloor}\{t\}^n \ dt+\int_0^{\{x\}}\{t\}^n\ dt\\ &=\lfloor x\rfloor\int_0^1t^n \ dt +\int_0^{\{x\}}t^n\ dt\\ &=\frac{\lfloor x \rfloor+\{x\}^{n+1}}{n+1} \end{align}

Important fact:

The indefinite integral $\int f(x) \ dx$ is defined to be a function $F(x)$ such that $F'(x) = f(x)$. This will exist by the Fundamental Theorem of Calculus if and only if $f$ is a continuous function.

Hence, in this case, since $\{x\}^n$ is not a continuous function from $\mathbb R$ to $\mathbb R$, the indefinite integral is not defined in this region! In general, it isn't a good idea to think of the indefinite integral as the area under the curve.

However, due to the properties of this function, it is still integrable, and it is still possible to define a definite integral. In particular, since the function is continuous on $[n,\ n+1)$ for $n \in \mathbb Z$, it will have an indefinite integral (or an antiderivative) in each of these regions. Given $x \in \mathbb R$, it is possible to integrate this function from $0$ to $x$ to get the area under the curve as shown above. However, this will be a definite integral and not an indefinite one.

• Okay, I understand how you get that. Where did I go wrong in my process? – user3002473 Aug 30 '14 at 20:39
• I think the key cause is that in trying to integrate from $0$ to $x$, you have used $x$ as a dummy variable - i.e. $\int_0^x \{x\}^n dx$. As such, your $x$ is appearing in two different varieties, which is always gonna cause confusion! I'm not sure I see how you got that the integral should be $\lfloor x \rfloor\int{x^ndx} + \{x\}\int{x^ndx}$ – Mathmo123 Aug 30 '14 at 20:41
• OH! I finally realized exactly what you mean! Okay, awesome, that makes total sense to me know. Thanks for your help! :) – user3002473 Aug 30 '14 at 20:46
• See my edit - it contains an import distinction – Mathmo123 Aug 30 '14 at 20:52