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The integral $\int _{\:0}^{\:\frac{\pi }{2}}\:1+\left(-sinx\right)+\left(-sinx\right)^2+...dx$ is a geometric sereis which can be expressed as $\int _0^{\frac{\:\pi }{\:2}}\:\frac{1}{1-\left(-sinx\right)}dx$.

But this is only valid if $|r|<1$. In the given bounds in the integral, $\sin{x}$ is precisely in between $0$ and $-1$, but at $\sin(90)$, it is actually equal to $-1$ which does not suit $|r|<1$

It always thought that the endpoints were inclusive in a definite integral but that seems to lead to contradictions here.

user71207
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  • What happens at any one particular point never matters for a definite integral (for an otherwise integrable function) – Ninad Munshi Feb 16 '21 at 08:55
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    $\sin (90)$ is $1$. Anyway, the value at the end points have no effect on the value of the definite integral. – Kavi Rama Murthy Feb 16 '21 at 08:55
  • This is an example of an improper integral. – Greg Martin Feb 16 '21 at 09:01
  • I don't understand - so is it included or not? It seems yes it is included by it has "no effect on the value", so it's not included at the same time?... – user71207 Feb 16 '21 at 09:08
  • @user71207 as the integrand is not bounded you have to integrate the function from $[0,\varepsilon]$ and take limit of $ \varepsilon \rightarrow \pi/2 $ – Aditya Dwivedi Feb 16 '21 at 09:09
  • And if the function was bounded then you can ignore the value at the end point – Aditya Dwivedi Feb 16 '21 at 09:10
  • The question is meaningless and I'll give an example why we phrased it that way because it's true for any point not just the endpoints. Consider $f(x) = 3$ and $$g(x) =\begin{cases} f(x) & x\neq 1\ 5 & x=1\end{cases}$$ Both of the integrals $$\int_0^2 f(x)dx = \int_0^2 g(x)dx = 6$$ even though the second one "excludes" the point $x=1$. If the integrals take on the same value whether or not any single point is "included", then inclusion is a bit of a meaningless thing to discuss. – Ninad Munshi Feb 16 '21 at 09:13
  • @NinadMunshi In the definition of integration we require the function to be bounded otherwise supremum and infimum won't make sense which should be the primary concern of this question – Aditya Dwivedi Feb 16 '21 at 09:15
  • @AdityaDwivedi Boundedness is not required for integrability. – Ninad Munshi Feb 16 '21 at 09:16
  • @NinadMunshi atleast you should look at the definition of the integral, what will you say the supremum of the function if it is not bounded. I think you meant that we can define an improper integral – Aditya Dwivedi Feb 16 '21 at 09:17
  • @AdityaDwivedi as a very simple counterexample, could you tell me what $$\int_0^1 \frac{dx}{\sqrt{x}}$$ evaluates to? – Ninad Munshi Feb 16 '21 at 09:18
  • @NinadMunshi That can be evaluated by integrating the function in $[\varepsilon,1]$ and take limits.And that is not a counterexample, if still not satisfied see this – Aditya Dwivedi Feb 16 '21 at 09:19
  • @AdityaDwivedi you are probably referring to Darboux integrabilitiy with the sup and inf. Boundedness is not a requirement, you can apply the same definition to $\frac{1}{\sqrt{x}}$ and prove it is integrable by restricitng the refinements on the partitions to never touch $0$. This does not have to be a limiting process as you describe. Anyway, feel free to open a chat window, I don't want to clutter OP's post any further to clear up misconceptions. – Ninad Munshi Feb 16 '21 at 09:24
  • Let us continue this discussion in chat. – Aditya Dwivedi Feb 16 '21 at 09:24
  • $$\int_{0}^{\pi /2}\left[1+\left(-\sin x\right)+\left(-\sin x\right)^2+\ldots ,\right]dx=$$ $$=\frac{\pi}{2}+\sum_{n=1}^{\infty}\int_{0}^{\pi /2}(-\sin x)^n,dx=\sum _{n=1}^{\infty } \frac{\sqrt{\pi } (-1)^n \Gamma \left(\frac{n+1}{2}\right)}{n \Gamma \left(\frac{n}{2}\right)}+\frac{\pi }{2}=1$$ – Raffaele Feb 16 '21 at 12:45

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