Question

How would you find the sum of this series?

Answer 1

\[\sum_{n=-\infty}^{+\infty}\frac{\sin(n)}{n}\]

I thought it would be better to make another question but with a specific series. I realize there may be multiple ways of finding this sum, but I'm mainly curious about a contor integration way, if one is known by anyone.

Answer 2

1) Does it converge?
2) Does it have a sum?

Answer 3

Its sum is \(\pi\). I just don't know how I would get to that conclusion. I may be asking too much that there be a contour integration way to show it, but I was hoping someone might know :)

Answer 4

Hmm, I keep making some mistake and ending up with \(-\dfrac{1}{2}\)...

Answer 5

I would assume you have to treat it like
\[1 + 2\sum_{n=1}^{+\infty}\frac{ \sin(n) }{ n }\]

I don't know if that changes the process of evaluating it if you have it in that form, though.

Answer 6

Here's what I have so far:

Since \(\dfrac{\sin n}{n}\) is even, you can consider a contour integral with \(f(z)=\dfrac{\sin z\cot\pi z}{z}\) with a square contour that surrounds the poles. You have poles at all the integers, which should be easy to deal with. I forget the details, but the integral \(\displaystyle \oint_C f(z)\,\mathrm{d}z\) should disappear, so you have
\[0=\sum_{k\in\mathbb{Z}}\text{Res}(f(z),k)=\text{Res}(f(z),0)+\sum_{k\in\mathbb{Z}\setminus\{0\}}\text{Res}(f(z),k)\](I only split the poles up because there's nice pattern with the non-zero poles' resiudes.)

Using the expansion is one way to compute the residues, but since the poles are simple, I'll just compute via the limit route:
\[\text{Res}(f(z),0)=\lim_{z\to0}z\frac{\sin z\cot \pi z}{z}=\dfrac{1}{\pi}\]For the rest, you have
\[\sum_{k\in\mathbb{Z}\setminus\{0\}}\lim_{\substack{z\to k\\k\in\mathbb{Z}}}(z-k)\frac{\sin z\cot \pi z}{z}=\sum_{k\in\mathbb{Z}\setminus\{0\}}\frac{\sin k}{k\pi }=\frac{2}{\pi}\sum_{k=1}^\infty\frac{\sin k}{k}\]where the last equality holds because \(\dfrac{\sin n}{n}\) is even, and so
\[\begin{align*}0&=\text{Res}(f(z),0)+\sum_{k\in\mathbb{Z}\setminus\{0\}}\text{Res}(f(z),k)\\[1ex]
-\frac{1}{\pi}&=\frac{2}{\pi}\sum_{k=1}^\infty\frac{\sin n}{n}\end{align*}\]I clearly have made some mistake somewhere, but I'm not sure where.

Answer 7

The residues should be fine, I've been checking back and forth with Mathematica.

Answer 8

Well, I know you need an extra \(\pi\) if the integrals are of that form. \(f(z)\pi \cot(\pi z)\) where our f(z) would be \(\frac{\sin(z)}{z}\), which should make that result come out to be just 1.

Answer 9

Hmm, I was under the impression that the inclusion of \(\pi\) in \(f(z)\) isn't necessary, as it should cancel out... At least that's what happened when I reviewed the Basel problem via contour integration from my old lecture notes.

Answer 10

Am I missing a pole somewhere?

Answer 11

Ah. Well, we were only given two different types of contour integrals, both of which used an extra \(\pi\). But I'm not sure if that changes with series of sine and cosine, we weren't given anything too specific on that, just that it was possible.

Answer 12

I initially thought the idea was to use exponential form, but doing that gave me problems with the contour. I was trying to make sure everything vanished, or at least gave a convergent result, but I was unable to do that.

Answer 13

I'm not familiar with the exponential form (though I doubt it changes much at first glance). I remember coming across a pdf on the web that surveyed only three variations of \(f(z)\): \(\cot\pi z\) for even summands, \(\csc\pi z\) for alternating ones, and \(\sec\pi z\) for odd.

Answer 14

Hm. We were given \(\cot(\pi z)\) and \(\csc(\pi z)\). Good to know there's a \(\sec(\pi z)\) one.

As for the exponential form idea, I guess it goes along with how you would integrate functions with sine and cosine. Like to integrate \(\frac{\sin(x)}{x}\) you would look at \(\frac{e^{iz}}{z} \), that's why I had that idea.

Answer 15

Ah here's the file I mentioned. Cached link: http://webcache.googleusercontent.com/search?q=cache:I7_GuQ5MlhAJ:www2.mae.ufl.edu/~uhk/SERIESCOMP.pdf+&cd=10&hl=en&ct=clnk&gl=us

Answer 16

So when I tried that idea, I was considering
\[\oint \frac{e^{iz}\pi \csc(\pi z)}{z} dz\]

And okay, Ill take a look :)

Answer 17

You would think that it'd be possible to find exact sums of things like \[\sum_{n=1}^{\infty}\frac{1}{n^{3}}\]
since \(\sec(\pi z)\) handles some odd functions. Oh well.

So you used \(\cot(\pi z)\), which is the idea for even functions. I suppose my thought process was that since sine and cosine alternate that \(\csc(\pi z)\) would work. And maybe it does in the general case, but the evenness of the function takes priority?

Answer 18

It seems to me that the choice of \(\cot\pi z\) is a natural one if \(f(n)\) is even, since the residues would be the same for positive/negative non-zero integers, so much so that it should work regardless of what \(f(n)\) is so long as it's even.

Answer 19

Yeah, and after messing around with it, it seems using \(\csc(\pi z)\) gives a \((-1)^{k}\) when computing residues, so it does no good unless your series truly is alternating.

Well, going through my notes, there's a small mention that \(f(z)\) is meromorphic without integer poles, which could be what is messing things up.

Answer 20

So maybe what could be done instead is to consider some other sum.
\[\sum_{n=-\infty}^{+\infty}\frac{ \sin(n) }{ n+\alpha }\] where \(\alpha \neq n\)

There may be a better sum to consider, but its an idea. If we have the sum of this general series, then we just let \(\alpha =0 \) in the final result.

Answer 21

I mean \(\alpha \neq -n \)

Answer 22

Turns out the problem is with the contour (or possibly the integrand).

We've been using the square centered at the origin with vertices at \(\left(\pm\left(N+\frac{1}{2}\right),\pm\left(N+\frac{1}{2}\right)\right)\).

We have to have\[\left|\oint_\gamma f(z)\,\mathrm{d}z\right|\le M\,\ell(\gamma)\]where \(M\) is some constant and \(\ell(\gamma)\) is the arc length of the contour. Obviously, \(\ell(\gamma)=8N+4\). However, our \(f(z)\) is not bounded in absolute value.

For instance, in the case where \(z=x+i y\) with \(x=N+\dfrac{1}{2}\) and \(|y|<\dfrac{1}{2}\) (the right edge of the contour), we have
\[\left|\oint_\gamma f(z)\,\mathrm{d}z\right|\le\oint_\gamma|f(z)|\,|\mathrm{d}z|=\frac{\pi(8N+4)\tanh\frac{\pi}{2}}{\sqrt{(N+\frac{1}{2})^2+y^2}}\to8\neq0\]as \(N\to\infty\).

Answer 23

Oops, didn't even realize this had a response to it, my apologies.

But yeah, ML Inequality, eh? So it would've never converged on that contour in the first place, got it. I'm glad you came up with that, it's definitely something I want to keep in mind for the future.

Don't suppose you came up with an appropriate contour, did ya?

Answer 24

Nothing comes to mind, but I think you can use the same one if you manipulate the integrand or summand. It's the \(\dfrac{1}{z}\) factor that's making this not work, I think. Integration/summation by parts could provide a way around it.