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I have really forgotten how to do this kind of calculation; would any one please tell me?

  1. $\int_0^{\pi/4} \lfloor3 \tan^2x\rfloor, dx$
  2. $\int_{-100}^{100} \lfloor t^3\rfloor,dt$
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    Oh, now I understood that you meant the greatest integer $f$unction.2012-03-26

2 Answers 2

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Let's start with 1. If $0\le x\le\pi/4$, then $0\le3\,\tan^2x\le3$. $ \lfloor3\,\tan^2x\rfloor=\begin{cases} 0 & \text{if } 0\le\,\tan^2x<1/3,\\ 1 & \text{if } 1/3\le\tan^2x<2/3,\\ 2 & \text{if } 2/3\le\tan^2x<1,\\ 3 & \text{if } \tan^2x=1. \end{cases} $ Let $ x_1=\arctan\sqrt{\frac{1}{3}},\quad x_2=\arctan\sqrt{\frac{2}{3}},\quad x_3=\arctan1. $ Then $ \int_0^{\pi/4}\lfloor3\,\tan^2x\rfloor\,dx=0\cdot\int_0^{x_1}dx+1\cdot\int_{x_1}^{x_2}dx+2\cdot\int_{x_2}^{x_3}dx=2\,x_3-x_2-x_1. $ I leave to you to find explicit values for $x_k$, $1\le k\le3$, and the extension of this method to problem 2.

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Question 1: Let $f(x)=\lfloor 3\tan^2 x\rfloor$. Look first at those $x$ such that $0\le 3\tan^2 x<1$. There $f(x)=0$. Note that $\arctan(1/\sqrt{3})=\pi/6$, so $f(x)=0$ for $0\le x<\pi/6$.

In the same way, if $1\le 3\tan^2 x <2$, then $f(x)=1$, and if $2\le 3\tan^2 x<3$ then $f(x)=2$.

So $f(x)=1$ for $\pi/6\le x<\arctan(\sqrt{2}/\sqrt{3})$, and $f(x)=2$ for $\arctan(\sqrt{2}/\sqrt{3})\le x<\pi/4$.

Unfortunately, $\arctan(\sqrt{2}/\sqrt{3})$ is I believe nothing familiar. Call it $a$. The calculator says that $a$ is approximately $0.6847192$.

Then our integral is $(\pi/6-0)(0)+(a-\pi/6)(1)+(\pi/4-a)(2)$. One could simplify this expression to $\pi/3-a$, and then compute.

Question 2: We use a trick to cut down on the work. Note that if $y$ is positive and is not an integer, then $\lfloor -y\rfloor=-\lfloor y\rfloor-1$. (This is just the observation that, for example, $\lfloor \pi \rfloor=3$ while $\lfloor -\pi\rfloor =-4$.)

For the integral from $-100$ to $0$, let $u=-x$. We have then $\int_{-100}^0 \lfloor x^3\rfloor \,dx=\int_{100}^0 \lfloor -u^3\rfloor (-du).$ For integration, changing the value of a function at a finite number of places makes no difference. When $u^3$ is not an integer, $\lfloor -u^3\rfloor=-\lfloor u^3\rfloor-1$, and therefore $\int_{100}^0 \lfloor -u^3\rfloor (-du)=\int_0^{100} (-\lfloor u^3\rfloor-1)du.$ So the full integral is equal to $\int_0^{100}\lfloor x^3\rfloor\,dx+ \int_0^{100} (-\lfloor u^3\rfloor-1)\,du.$ In the second integral, change the dummy variable of integration to $x$. There is very nice cancellation, and we end up with $\int_0^{100} (-1)\,dx.$

Remark: Question $2$ turns out to be easy. (Note that finding the integral from $0$ to $100$ is far more unpleasant.) But I should confess that I first did it by breaking up the region (at first wrongly and then correctly) into intervals, and observing the almost cancellations. Writing up that version would be a nuisance, hence the simpler version. It is, however, not a bad idea to begin to do it the grungy way.

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    Julián Aguirre: Thanks, I was quite wrong.2012-03-26