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(1) Evaluation of \(\displaystyle \int_{0}^{\pi}\lfloor \cot x \rfloor dx\;,\) where \(\lfloor x \rfloor \) is floor function of \(x\).

(2) Evaluation of \(\displaystyle \int_{0}^{\pi}\lfloor \cos x \rfloor dx\;,\) where \(\lfloor x \rfloor \) is floor function of \(x\).

My Try for (1):: Let \(\displaystyle I = \int_{0}^{\pi}\lfloor \cot x \rfloor dx\;,\) Now using \(\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx\)

So we get \(\displaystyle I = \int_{0}^{\pi}\lfloor -\cot x \rfloor dx\;,\) So we get

\(\displaystyle I = \int_{0}^{\pi}\left(\lfloor \cot x \rfloor + \lfloor -\cot x \rfloor \right)dx = -\int_{0}^{\pi}dx = -\pi\)

So We get \(\displaystyle I = \int_{0}^{\pi}\lfloor \cot x \rfloor dx = -\frac{\pi}{2}\)

Here above we have used the property \(\lfloor x \rfloor +\lfloor -x \rfloor = -1\;,\) When \(x\notin \mathbb{Z}\).

actually i have a confusion can we write same for \(\lfloor \cot x \rfloor + \lfloor -\cot x \rfloor = -1\;,\) when \(\cot x\notin \mathbb{Z}\)

For (2) I have solve it using drawing of graph of \(y = \lfloor \cos x \rfloor \) in \(x\in \left(0,\pi\right)\)

which is \(\displaystyle y=\lfloor \cos x \rfloor = 0\;,\) when \(x\in \left(0,\frac{\pi}{2}\right]\) and \(\displaystyle y=\lfloor \cos x \rfloor = -1\;,\) when \(x\in \left(\frac{\pi}{2},\pi\right)\)

Jacks, the first integral diverges. Mathematica also agrees with this. Besides, I had drawn a graph of that function on Geogebra which confirms that. So, how were you able to evaluate it?

Indeed as Apostolos says, the integral diverges. What jacks's proof shows is that for every \(0 < \varepsilon < \pi/2\) we have that \(\int_{\varepsilon}^{\pi-\varepsilon} \lfloor \cot{x} \rfloor \; dx = -\pi + 2\varepsilon.\) However this does not mean that the value of the integral converges to\(−\pi\). The integral tends to \(+\infty\) near \(0\) and to \(-\infty\) near \(\pi\) and we are not allowed to cancel them out.

Sometimes we might say that the Cauchy principal value of this integral is \(-\pi\).

After Demetres remark, here is my solution:

$$\begin{aligned}
\int_{0}^{\pi}\left \lfloor \cot x \right \rfloor\,dx &\overset{u=\cot x}{=\! =\! =\! =\!}-\int_{-\infty}^{\infty}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du \\
&= -\left ( \int_{-\infty}^{0}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du +\int_{0}^{\infty}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du\right )\\
&= -\left ( I+J \right )
\end{aligned}$$

Then:
$$\begin{aligned}
J &=\int_{0}^{\infty}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du \\
&=\int_{0}^{1}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du + \int_{1}^{2}\frac{\left \lfloor u \right \rfloor}{1+u^2}\,du+\cdots \\
&= \sum_{n=1}^{\infty}n\int_{n}^{n+1}\frac{du}{1+u^2}=\sum_{n=1}^{\infty}n\left [ \tan^{-1}(n+1)-\tan^{-1}(n) \right ]\geq \sum_{n=1}^{\infty}n=+\infty
\end{aligned}$$

Thus \(J\) diverges and so does the initial integral.


Jacks, I would like to add an additional question that arose from the exercise and my first wrong solution and it is not that hard.
Evaluate the series: \( \displaystyle \sum_{n=1}^{\infty}\left ( \tan^{-1}\left ( n+1 \right )-\tan^{-1} (n) \right ) \).

Tolaso J Kos wrote:Jacks, I would like to add an additional question that arose from the exercise and my first wrong solution and it is not that hard.
Evaluate the series: \( \displaystyle \sum_{n=1}^{\infty}\left ( \tan^{-1}\left ( n+1 \right )-\tan^{-1} (n) \right ) \).

We define the sequence \(\displaystyle{\left(a_{n}\right)_{n\in\mathbb{N}}}\) by

\(\displaystyle{a_n=\sum_{k=1}^{n}\left(\arctan\,(k+1)-\arctan\,k\right)\,, n\in\mathbb{N}}\).

Now we see that \(\displaystyle{a_1=\arctan\,2-\arctan\,1=\arctan\,2-\frac{\pi}{4}}\).

Let \(\displaystyle{a_n=\arctan\,(n+1)-\frac{\pi}{4}\,\,(\ast)}\) , for some \(\displaystyle{n\in\mathbb{N}}\) .

\(\displaystyle{n\to n+1}\) :

\(\displaystyle{\begin{aligned} a_{n+1}&=a_n+\arctan\,(n+2)-\arctan\,(n+1)\\&\stackrel{(\ast)}{=}\arctan\,(n+1)-\frac{\pi}{4}+\arctan\,(n+2)-\arctan\,(n+1)\\&=\arctan (n+2)-\frac{\pi}{4}\end{aligned}}\)

So, \(\displaystyle{a_{n}=\arctan\,(n+1)-\dfrac{\pi}{4}\,,n\in\mathbb{N}}\) .

Since, \(\displaystyle{\lim_{n\to \infty}\arctan\,(n+1)=\dfrac{\pi}{2}}\), we have that :

\(\displaystyle{\begin{aligned}\sum_{k=1}^{\infty}\left(\arctan\,(k+1)-\arctan\,k\right)&=\lim_{n\to \infty}a_{n}\\&=\lim_{n\to \infty}\left(\arctan\,(n+1)-\frac{\pi}{4}\right)\\&=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\pi}{4}\end{aligned}}\) .