Does $\int_0^{\pi \over 2} \lfloor \tan(x) \rfloor\, dx$ converge?

Question

My book follows the following method

Let

$$I=\int_0^{\pi \over 2} \lfloor \tan(x) \rfloor\, dx.$$

Then using King's rule $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$ we have

$$I=\int_0^{\pi \over 2} \lfloor \tan(\pi- x)\rfloor\, dx=\int_0^{\pi \over 2} \lfloor \tan(- x)\rfloor\, dx$$

Adding the above two we get:

$$2I =\int_0^{\pi\over 2} \left[\lfloor(\tan(- x)\rfloor+\lfloor\tan x\rfloor\right]\,dx$$

Now since $ \lfloor x\rfloor +\lfloor -x\rfloor=-1$ when $x$ is not a integer and $0$ otherwise the integral becomes

$$2I=\int_0^{\pi\over 2} -1\, dx$$

which then gives $I = {-\pi \over 2}$

But Wolfram Alpha says that the integral does not converge.

My Question:

Is my book correct? If not where is the error in above calculations?

Answer 1

Your application of King's rule is wrong; it gives$$I=\int_0^{\pi/2}\lfloor\tan(\color{blue}{\pi/2}-x)\rfloor dx.$$To prove $I$ diverges, note$$I\ge\int_0^{\pi/2}(\tan x-1)dx=[\ln|\sec x|-x]_0^{\pi/2}=\infty.$$

Answer 2

The problem with your reasoning is that you assume that $$\exists \int (f+g) \implies \exists \int f \land \exists \int g$$ which is not true.

To prove that it diverges, note that $$x=\lfloor x \rfloor + \{x\}$$ So $$\lfloor \tan(x) \rfloor=\tan(x)-\{\tan(x)\}\geqslant\tan(x)-1$$ And $$\int_0^{\pi/2}\tan(x)\mathrm{d}x$$ diverges.

Answer 3

$$I=\int_{0}^{\pi/2} [\tan x] dx=\int_{0}^{\infty} [t] \frac{dt}{1+t^2}$$ $$ \frac{t-1}{1+t^2}\le\frac{[t]}{1+t^2} \le \frac{t}{1+t^2}$$ $$ \implies [\frac{1}{2}\ln (1+t^2)-\tan^{-1} t]_{0}^{\infty} \le I \le \frac{1}{2} \ln(1+t^2)|_{0}^{\infty}.$$ As both the left and the right integral diverge so does $I$.