2

Find the value of $$\int_{0}^{x} \lfloor \cot(t)\rfloor\,dt$$ where $x \in [(4n+1)\frac {\pi}{2},(4n+3)\frac {\pi}{2}]$ and $n\in N $ Also $\lfloor\cdot \rfloor$ represnts the greatest integer function.

I tried this by graphical method but i am unable to obtain the general solution. I succeed only in to obtain the area in period. That is the signed area of given integrand in $ (0,\pi) $ is $-\frac {\pi}{2} $.

I am giving my solution . Please see the imageimage 1

image2

The exact answer of this question is $(2n+1)\frac {\pi}{2}-x$ Any one can explain. How to obtain it enter image description here

Girish Kumar Chandora
  • 931
  • 1
    The integral blows up no matter what $x$ is, since $\int_0^\epsilon \lfloor(\cot x) \rfloor = \infty$. There is a chance that the signed area will cancel out if $x$ is another pole, but be careful in assuming this exists – Brevan Ellefsen Aug 14 '17 at 21:37
  • 1
    If you meant inverse cotangent then your "exact answer" still isnt correct, but see here – Brevan Ellefsen Aug 14 '17 at 22:00
  • @BrevanEllefsen you are wrong. In my question it is cotangent not inverse cotangent. The answer is correct as you can see the image of my solution. If its not correct then explain the mistake in my solution for intereval o to pi – Girish Kumar Chandora Aug 15 '17 at 04:10
  • Well, if you do mean the cotangent then see my first comment. Your integral blows up at the poles $\pi n$, which includes the point $0$, so the integral from $0$ to ANYTHING is undefined (except perhaps integrals from one pole to another, such as $\int_0^\pi \lfloor \cot(x) dx \rfloor $). As to your diagram, I have no idea what is happening in it, but I do notice that you don't have any rectangles around the point $0$. That could be a big mistake. – Brevan Ellefsen Aug 15 '17 at 04:17
  • Wolfram Alpha calculation for $x=1$. Note that the same thing happens with any $x$ you choose. – Brevan Ellefsen Aug 15 '17 at 04:20
  • Wolfram Alpha Calculation for $x = \frac{5 \pi}{2}$, which is the case $n=1$ for your definition of $x$ – Brevan Ellefsen Aug 15 '17 at 04:21
  • Wolfram is 99% correct not always i can give u questions in which the wolfram answer will be wrong – Girish Kumar Chandora Aug 15 '17 at 04:21
  • @BrevanEllefsen at o it took the limit n tends to infinity. Please see the image of my solution. – Girish Kumar Chandora Aug 15 '17 at 04:22
  • Sure. But it's not. You haven't actually proven the integral converges yet... that is always something you should do with integrals like this. If you want, pull up a graph of $\lfloor \cot(x) \rfloor$ in Desmos.com or something. I don't wish to argue this any further, so don't expect a response until you actually investigate the function more. – Brevan Ellefsen Aug 15 '17 at 04:23
  • As another suggestion before I go, note that $\lfloor \cot(x) \rfloor \ge \cot(x)-1$ so that, assuming the integral exists, $\int_0^1 \lfloor \cot(x) \rfloor dx \ge \int_0^1 \cot(x) dx = +\infty$. You can modify this to prove divergence for any upper bound $x$ – Brevan Ellefsen Aug 15 '17 at 04:26
  • @BrevanEllefsen ok i will try my best to explain in detail. I also dont want to argue. The answer is correct – Girish Kumar Chandora Aug 15 '17 at 04:30
  • @BrevanEllefsen $\lfloor \cot(x) \rfloor \ge \cot(x)-1$ so that, assuming the integral exists, $\int_0^\pi \lfloor \cot(x) \rfloor dx \ge \int_0^\pi \cot(x) dx = ? . Why the integral of cotx from 0 to \pi is not zero . As we can see from graph by symmetry the area above and below x axis are in equal magnitude. Its integral will be 0 or not? – Girish Kumar Chandora Aug 15 '17 at 04:44
  • Well, that was exactly what I was saying earlier. The ONLY points where this integral even could converge in some sense is when both limits are at poles, i.e. we have $\int_0^{\pi n} f(x) dx$. In this case you have to actually confirm it exists, as you are no longer using a standard Riemann Integral. At best you have an improper Riemann Integral and you have to calculate $\lim_{\epsilon \to 0} \int_\epsilon^{\pi - \epsilon} \lfloor \cot(x) \rfloor dx$. Does that make sense? – Brevan Ellefsen Aug 15 '17 at 04:54
  • @BrevanEllefsen i asked the question which objective type and there are four options in terms of x ,n and /pi. – Girish Kumar Chandora Aug 15 '17 at 04:57
  • ?? I'm not sure what you are asking. Regardless, after doing some quick numerical tests it appears that $\lim_{\epsilon \to 0} \int_\epsilon^{\pi - \epsilon} \lfloor \cot(x) \rfloor dx = -\frac{\pi}{2}$ like you want, so that's good. The problem is that you haven't even specified if this is the integral you are wishing to use, or if you are using Riemann Integration, etc. – Brevan Ellefsen Aug 15 '17 at 04:58
  • @BrevanEllefsen you can see the original image of actual question. – Girish Kumar Chandora Aug 15 '17 at 05:01
  • Also, $\lim_{\epsilon \to 0} \int_\epsilon^{\pi - \epsilon} \cot(x)-1 dx = -\pi$ it would appear, so my bound above does still hold. – Brevan Ellefsen Aug 15 '17 at 05:01

1 Answers1

1

Let $k \in \mathbb Z$. Then for $0 < t < \pi$, the condition $\lfloor \cot t \rfloor = k$ implies $$\cot^{-1} k \ge t > \cot^{-1}(k+1).$$ Note the reversal of the direction of the inequality. It follows that the improper integral $$\begin{align*} \int_{t=0}^{\pi/2} \lfloor \cot t \rfloor \, dt &= \sum_{k=0}^\infty k \left(\cot^{-1} k - \cot^{-1}(k+1)\right) \\ &= \lim_{N \to \infty} \sum_{k=0}^{N-1} k \cot^{-1} k - \sum_{k=1}^N (k-1) \cot^{-1} k \\ &= \lim_{N \to \infty} \sum_{k=1}^{N-1} \cot^{-1} k - (N-1) \cot^{-1} N \\ &= -1 + \sum_{k=1}^\infty \cot^{-1} k \\ &> -1 + \int_{x=1}^\infty \cot^{-1} x \, dx \\ &= -1 + \left[x \cot^{-1} x + \frac{\log (1+x^2)}{2} \right]_{x=1}^\infty \end{align*} $$ and it is now clear that the value of the original integral cannot be bounded.

heropup
  • 135,869