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$$\int_{0}^2\frac{\sin2x}{2x^2-\pi x}dx$$

I simply have to find the convergence of this integral between [0,2]. First I was tasked with doing an integral decomposition, since this integral is undefined at x=0 and x= pi/2. The decomposition that the question wants looks like this: $$\int_{0}^1\frac{\sin2x}{2x^2-\pi x}dx+\int_{1}^\frac{\pi}{2}\frac{\sin2x}{2x^2-\pi x}dx+\int_{\frac{\pi}{2}}^2\frac{\sin2x}{2x^2-\pi x}dx$$

I need to determine if each of these integral converge or not. I have attempted using the Comparison Integral Theorem and I get stuck. I'm pretty confident they are all suppose to be convergent simply because on Desmos it looks convergent and putting in the integral to Desmos you get a finite number. Any help would be much appreciated. Just to note I am in first year university Calculus II and this is just the 1st month of the term so far.

Edit: I think I may be able to find the convergence of $$\int_{0}^2\frac{\sin2x}{2x^2-\pi x}dx$$

if I just find the limit of the function, and if the limit exists would that mean the integral is convergent? My question is now would this limit, $\lim_{x\to \infty}\frac{\sin2x}{2x^2-\pi x}$ prove that the integral, $\int_{0}^2\frac{\sin2x}{2x^2-\pi x}dx$, converges?

amWhy
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Spencer Ken
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    Why did you try to vandalize my answer? – José Carlos Santos Feb 08 '22 at 16:56
  • I feel like this question could be under copyright since it was a question given by profs at a university. But I also cannot delete it so idk – Spencer Ken Feb 08 '22 at 16:59
  • Questions take from published textbooks appear here all the time. What's so special about yours? – José Carlos Santos Feb 08 '22 at 17:01
  • It was most likely written by the profs, not from a textbook. I just don't want to get in trouble if they see this. – Spencer Ken Feb 08 '22 at 17:01
  • It is probably unlikely they will see it by the time the solutions for this assignment are posted, but I didn't want to take the risk. – Spencer Ken Feb 08 '22 at 17:03
  • I've had students of mine posting here questions from me, and I never saw anything wrong with that. – José Carlos Santos Feb 08 '22 at 17:05
  • Okay sorry, the one question would be worth a very small amount of it anyways. – Spencer Ken Feb 08 '22 at 17:09
  • @Spencer Is the question from a current exam??? What institution are you attending, and what is the profs name? – amWhy Feb 08 '22 at 17:17
  • I imagine, @JoséCarlosSantos, that you'd have problems if you encountered exam problems? Especially in Universities still using remote learning? – amWhy Feb 08 '22 at 17:18
  • It is not an exam problem. It is a homework assignment problem. – Spencer Ken Feb 08 '22 at 17:19
  • @amWhy Sure, but not copyright problems, which is what the OP mentioned in his first comment. Besides, did you notice that this questin was originally posted nine hours ago? – José Carlos Santos Feb 08 '22 at 17:20
  • @amWhy I do not think any universities would allow access to the internet during an exam, and my university uses Proctortrack, which records you and your entire screen. I was working on the question for hours tell late last night, so I just decided to ask it on here as I did not have many other options. – Spencer Ken Feb 08 '22 at 17:23
  • @amWhy This is not even the full question, there is a) b) and this is just c). And it is not the full question simply the function. I should have just asked if the function was integrable on the interval [0,2], not referred to it as from an assigned questions – Spencer Ken Feb 08 '22 at 17:24
  • @JoséCarlosSantos I see only edits from you on your answer, so how did the vandalization occur? – Тyma Gaidash Feb 12 '22 at 03:08
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    @TymaGaidash By editing the question in such a way that my answer is not an answer anymore. – José Carlos Santos Feb 12 '22 at 03:22

2 Answers2

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There are two points $x\in[0,2]$ at which the expression $\frac{\sin(2x)}{2x^2-\pi x}$ is undefined: $x=0$ and $x=\frac\pi2$. But$$\lim_{x\to0}\frac{\sin(2x)}{2x^2-\pi x}=\lim_{x\to0}\frac{\sin(2x)}{2x}\cdot\frac2{2x-\pi}=-\frac2\pi$$and a similar computation shows that $\lim_{x\to\pi/2}\frac{\sin(2x)}{2x^2-\pi x}=-\frac2\pi$ too. Therefore, if you define$$\begin{array}{rccc}f\colon&[0,2]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}-\frac2\pi&\text{ if }x=0\text{ or }x=\frac\pi2\\\frac{\sin(2x)}{2x^2-\pi x}&\text{ if }x\in(0,2]\setminus\left\{\frac\pi2\right\},\end{cases}\end{array}$$then$$\int_0^2\frac{\sin(2x)}{2x^2-\pi x}\,\mathrm dx=\int_0^2f(x)\,\mathrm dx$$and $f$ is integrable, since it is continuous.

José Carlos Santos
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  • Is the function still continuous between 0 and 2? The reason I ask is because I thought the function had to be defined at all points to show continuity but it is not defined at 0 or pi/2. Can we still say it is continuous since the limits as the function approaches those points exist? – Spencer Ken Feb 08 '22 at 17:14
  • My function $f$ is defined at every point of $[0,2]$ and it is continuous there. – José Carlos Santos Feb 08 '22 at 17:21
  • So we can define a function f, that is the exact same as the function I asked about except that it is continuous on every point [0,2]. Sorry I am a little confused. – Spencer Ken Feb 08 '22 at 17:28
  • Yes, that is what I did. – José Carlos Santos Feb 08 '22 at 17:31
  • Okay, interesting. So since the two functions are exactly the same, if the integral of one is convergent then the integral of the other must also be convergent, right? – Spencer Ken Feb 08 '22 at 17:36
  • No. They are not exactly the same. The domain of one of them is $(0,2]\setminus\left{\frac\pi2\right}$, whereas the domain of the other one is $[0,2]$. But, since the second one is integrable, the fist one is integrable too. – José Carlos Santos Feb 08 '22 at 17:40
  • Why is that true that since the first one is integrable so is the second? – Spencer Ken Feb 08 '22 at 17:45
  • The meaning of $\int_0^1\frac{\sin(2x)}{2x^2-\pi x},\mathrm dx$ is $\lim_{a\to0}\int_a^1\frac{\sin(2x)}{2x^2-\pi x},\mathrm dx$. But$$\lim_{a\to0}\int_a^1\frac{\sin(2x)}{2x^2-\pi x},\mathrm dx=\lim_{a\to1}\int_a^1f(x),\mathrm dx=\int_0^1f(x),\mathrm dx.$$The same approach explains why is it that there is not problem with $\frac\pi2$. – José Carlos Santos Feb 08 '22 at 17:50
  • So we can say there exists a function that has the same behaviour as $\int_{0}^2\frac{\sin2x}{2x^2-\pi x}dx$. But is continuous on [0,2], without actually finding that function? – Spencer Ken Feb 08 '22 at 17:57
  • Well, we can say that. There is no law against it. But that is not what I said. I proved that such a function exists by actually describing it. – José Carlos Santos Feb 08 '22 at 18:01
  • By describing the behaviour in the interval, that proves such a function exists? – Spencer Ken Feb 08 '22 at 18:07
  • The function $f$ exists because I have described $f(x)$ for every $x\in[0,2]$. – José Carlos Santos Feb 08 '22 at 18:08
  • Okay thank you for answering my questions I really appreciate it. – Spencer Ken Feb 08 '22 at 18:09
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Clearly, $f(x)$ is continuos when $x\to 2^-$.

In $I^+(0)$ we may have some problems. Noticing that the denominator is asymptotic to $-\pi x$ because $2x^2 = o(x)\,\, x\to 0$ and using the fact that $\sin(x)\,\, \sim\,\, x \,\, x\to 0$: $$\lim_{x\to 0}\frac{\sin(x)}{-\pi x+o(x)}\,\,\sim\,\,\lim_{x\to 0}\frac{2x}{-\pi x}=-\frac{2}{\pi}$$ So, $f(x)$ is extensible with continuity in $I(0)$.

Also, $2x^2-\pi x=x\cdot(2x-\pi)=0 \implies x=\frac{\pi}{2}$. We know that $\sin\left(2\cdot\frac{\pi}{2}\right)=0$, so: $$\lim_{x\to \frac{\pi}{2}}\frac{\sin(2x-\pi+\pi)}{2x^2-\pi x}\,\,\sim\,\, \lim_{x\to \frac{\pi}{2}}\frac{x-\frac{\pi}{2}}{2x\cdot\left (x-\frac{\pi}{2}\right)}\,\,\sim\,\, \lim_{x\to \frac{\pi}{2}}\frac{x-\frac{\pi}{2}}{\pi}\cdot\frac{1}{\left (x-\frac{\pi}{2}\right)}=\frac{1}{\pi}$$ So, $f(x)$ is extensible with conituinity in $x=\frac{\pi}{2}$. Thus, it's integrable in generalised sense.

Matteo
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