# Antiderivative of $\lfloor\frac{2}{1+x^2}\rfloor$

I was supposed to find the definite integral $$\int_{0}^{\infty}\lfloor\frac{2}{1+x^2}\rfloor\mathrm dx$$ which can done easily by sketching the graph of $$y=\frac{2}{1+x^2}$$ and comes out to be $$1$$.

$$\int\lfloor\frac{2}{1+x^2}\rfloor\mathrm dx$$ where $$[.]$$ represents the floor function

I was wondering if there is a closed form anti-derivative for the function, by writing the integral as sum of integrals on intervals where the floor function can be removed from the integrand. And if there isn't a closed form, should this be written as a piecewise function with the values in its range being the possible definite integral values it can give on certain intervals. I have no idea how to proceed. Any hints would be appreciated. Thanks

• I haven't covered it yet, but this function isn't uniformly continuous, is it? Reading your problem, this came to my mind. Does it change anything? Mar 6, 2020 at 6:34
• Well. $[\frac 2{1+x^2}]=0$ for all $x< -1$ so $F(x)=0$ for $x < =1$. And $[\frac 2{1+x^2}]=1$ for $-1<x < 0$ so $F(x)= 1*(x-(-1))=x+1$ for $-1\le x< 0$. $[\frac 2{1+x^2}]=2$ for $x=0$ but that's a single point with no measure so $F(x)=x+1$ for $-1\le x\le 0$. For $0< x< 1$ so $F(x)=1*(x+1)$ for $0< x \ge 1$ and $[\frac 2{1+x^2}]=0$ for $x>1$ so for $x > 1$ then $F(x)=F(1)=2$. So $F(x)=\begin{cases} 0&x<-1\\x+1&-1\le x\le 1\\2&x> 1\end{cases}$. Mar 6, 2020 at 6:35

You sketched the graph of $$f$$ where $$f(x)=\left\lfloor\frac{2}{1+x^2}\right\rfloor$$. So you can sketch a graph of $$F(x)=\int_0^x \left\lfloor\frac{2}{1+t^2}\right\rfloor\, dt$$. You will see that it is piecewise linear, sort of like ___/--- if you will allow the ASCII art approximation. One way to express that function without using piecewise function notation happens to be: $$F(x)=\frac{\lvert x+1\rvert-\lvert x-1\rvert}2=\frac{2x}{\lvert x+1\rvert+\lvert x-1\rvert}$$
This is not really an "antiderivative". The derivative of $$F$$ is not $$f$$, because of the semi-continuous behaviour of $$f$$ at $$-1$$, $$0$$, and $$1$$. The differences are that (a) $$F'(0)=1$$, whereas $$f(0)=2$$. And (b), $$F'(-1)$$ and $$F'(1)$$ are undefined, where $$f(-1)=f(1)=1$$. But aside from those three places, $$F'(x)=f(x)$$.
In fact you can't have a true antiderivative of $$f$$. That is you cannot have a differentiable-everywhere function $$F$$ such that $$F'=f$$. Because if $$F$$ is differentiable, then $$F$$ is continuous. $$F$$ must be linear with slope $$1$$ in a punctured neighborhood of $$0$$, so $$F'(0)$$ must equal $$1$$. But $$f(0)$$ is $$2$$. So it's not really possible.
• Thanks for answering. I've plotted the graph of $\int_{0}^{x}\lfloor \frac{2}{1+t^2}\rfloor \mathrm dt$ on Desmos, but it only seems to be defined within $[-0.7,+0.7]$, that is not clear. It should be for the entire interval from $[-1,+1]$. Right? Mar 6, 2020 at 7:42
• I don't know, a Desmos bug? It seems to be plotting over $[-1/\sqrt{2},1/\sqrt{2}]$ for some reason. Desmos is good for simple things, but not good for everything. In GeoGebra, enter one line as A=(c,NIntegral(floor(2/(1+x^2)),0,c)) and it will create a slider for c, and you can move it to see the graph. Turn on trace for the point $A$ if you like. Mar 6, 2020 at 8:23
• In Desmos, you can also enter the formula I gave for $F$, and then ask it to plot $F'$. Then compare that to what it plots for $f$. Mar 6, 2020 at 8:28