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I am trying to calculate the definitive integral by definition (with Riemann sum). $$\int_{\frac{-\pi}{2}}^{\frac{3\pi}{2}} (2\sin{(2x+\frac{3\pi}{2})}) \ dx$$

But during the process of calculating it I have troubles evaluating this finite sum:

$$\sum_{i=1}^{n} \sin({\frac{\pi}{2}+\frac{4 \pi i}{n}})$$

I would greatly appreciate if you could help me figure out a nice expression for this using high school techniques.

Jesus
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    Do you have a good reason to believe that there is a nice way to evaluate this sum? – Physical Mathematics Apr 18 '20 at 18:36
  • Well, this question was given to me by a professor so unless he made a mistake I think there should be a way to evaluate it. Also, do you have any non-nice way to evaluate it in mind, although I'm afraid I wouldn't understand it? – Jesus Apr 18 '20 at 18:37
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    No I don't, it just easier to evaluate the integral by antidifferentiation. It is not always the case you can do it (in a reasonable way) using Riemann sums directly like this. – Physical Mathematics Apr 18 '20 at 18:39
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    I understand, but the question was "do it by definition", and even though I'm not 100% sure they meant Riemann sums I think so because in our book they were defined by Riemann sums. – Jesus Apr 18 '20 at 18:41
  • Often the antiderivative gives some direction for performing the summation. – copper.hat Apr 18 '20 at 18:46
  • $\sin(\frac{\pi}{2} + x) = \cos x$, and then this will help: https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro?rq=1 – Vishu Apr 18 '20 at 18:49
  • Are you familiar with Euler's formula? – copper.hat Apr 18 '20 at 19:01
  • I have read a bit about Euler's formula but I only know the basics. Maybe you should give it a try and then I'll ask you if I don't understand anything? – Jesus Apr 18 '20 at 19:04
  • There is a simpler approach, see @user775214's answer. Basically deal with $n$ even/odd separately and show that the 'second' half of the sum exactly cancels the first half. Much simpler than my apporach. – copper.hat Apr 18 '20 at 19:25
  • You can fool around with the formulas for sine of angle sums, extend them to a formula for $\sin n \alpha$ and add those up, but that looks very messy. – vonbrand Apr 18 '20 at 22:10

2 Answers2

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Consider the set of points

$$[\sin(\pi/2 + 4 \pi i/n),\cos(\pi/2 + 4 \pi i /n)] \subset \mathbf{R}^2, \ i = 1,\ldots n.$$

if $n$ is odd, these are the vertices of a regular $n$-gon centered at the origin, and if $n$ is even, they are the vertices of a regulat $n/2$-gon centered at the origin which each point occuring with multiplicity two.

In either case, the set of points is invariant by the rotation by $4 \pi/n$. Thus the average of the points is also invariant by this rotation. But the only point invariant by a non-trivial rotation is zero. Hence, if $n \ne 1,2$, the sum of all the points is zero, and thus the sum is zero for $n > 2$.

user775214
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Euler's formula is $e^{it} = \cos t + i \sin t$ so $\sin t = \operatorname{im} e^{it}$.

The Riemann sum with a uniform partition is ${2 \over n} \sum_{k=0}^{n-1} \sin (-{\pi \over 2}+ 2 \pi {k \over n} ) = {2 \over n} \operatorname{im}\left( \sum_{k=0}^{n-1} e^{i(-{\pi \over 2}+ 2 \pi {k \over n} )} \right)$. The advantage of this is the summation becomes a geometric series.

\begin{eqnarray} \sum_{k=0}^{n-1} e^{i(-{\pi \over 2}+ 2 \pi {k \over n} )} &=& e^{-i{\pi \over 2}} \sum_{k=0}^{n-1} (e^{i {2 \pi \over n} })^k \\ &=& e^{-i{\pi \over 2}} { (e^{i {2 \pi \over n} })^n -1 \over e^{i {2 \pi \over n} } -1 } \\ &=& 0 \end{eqnarray} If you do not want to use the Euler formula, you could note that if you split the sum into two appropriate parts, the second part cancels the first exactly (a glance at a plot will show this). You need to deal with odd & even $n$, but it is straightforward. (This is what the other answer is getting at, albeit in a more complicated manner.)

copper.hat
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