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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

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  • $\begingroup$ 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? $\endgroup$
    – Invisible
    Mar 6, 2020 at 6:34
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    $\begingroup$ 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 $F(x)=\begin{cases} 0&x<-1\\x+1&-1\le x\le 1\\2&x> 1\end{cases}$. $\endgroup$
    – fleablood
    Mar 6, 2020 at 6:35

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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.

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  • $\begingroup$ 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? $\endgroup$ Mar 6, 2020 at 7:42
  • $\begingroup$ 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. $\endgroup$
    – 2'5 9'2
    Mar 6, 2020 at 8:23
  • $\begingroup$ 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 $f$. $\endgroup$
    – 2'5 9'2
    Mar 6, 2020 at 8:28
  • $\begingroup$ That is piecewise, disguised... $\endgroup$
    – vonbrand
    Mar 6, 2020 at 9:56
  • $\begingroup$ @vonbrand Maybe, but there is no claim otherwise. Just "without using piecewise function notation". $\endgroup$
    – 2'5 9'2
    Mar 6, 2020 at 15:53

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