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This is a simple question, but its been bugging me. Define the function $\gamma$ on $\mathbb{R}\backslash \mathbb{Z}$ by $$\gamma(x):=\sum_{i\in \mathbb{Z}}\frac{1}{(x+i)^2}$$ The sum converges absolutely because it behaves roughly like $\sum_{i>0}i^{-2}$.

Some quick facts:

  • Pretty much by construction, $\gamma$ is periodic with period $1$.
  • As it approaches any integer from the left or right, it goes to positive infinity.
  • It is symmetric at every half integer; that is, $\gamma(n+1/2+x)=\gamma(n+1/2-x)$ for all $n\in \mathbb{Z}$ and $x\in \mathbb{R}$.

Can $\gamma$ be expressed in terms of more familiar (presumably trigonometric) functions?

My best guess is $\gamma(x)=\sin^{-2}(\pi x)$, but this is based more on what I hope it would be, rather than what it is.

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3  
Not a bad guess! You're off by a factor of $\pi^2$. –  Michael Lugo Jan 6 '10 at 20:39
    
I just added the complex variables tag, a little late, because all of the answers so far involve complex analysis in some way. In the case of David Speyer's answer this isn't explicit, but the product expansion of sine comes from an application of Weierstrass's factorization theorem. (Although I would be interested to know of other ways to get the sine product.) –  Jonas Meyer Feb 20 '10 at 22:41
    
Oh, answering my own implied question, it looks like the product formula can be derived using the gamma function on the reals, without need of complex numbers: ocw.mit.edu/NR/rdonlyres/728D634B-77C0-4B9A-A94C-28854771185E/0/…. Hopefully the tag is appropriate nonetheless. –  Jonas Meyer Feb 20 '10 at 22:56

4 Answers 4

up vote 15 down vote accepted

Use residues! For an entertaining narrative with the correct answer, see here. For a derivation, see a complex analysis text.

Added

Here's a link to an outline of the residue method of solving this: Conway page 122.

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Another proof:

Start with the identity

$$\sin (\pi x)= \pi x \prod \left( 1-x^2/i^2 \right)$$

and apply $\(d/dx)^2 \log$ to both sides.

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If you consider the two-dimensional version of your sum you are quickly led to the Weierstrass elliptic functions:

Naively, one is led to consider $$\sum_{m,n\in \mathbb{Z}}\frac{1}{(z+m+\omega n)^2}$$ (where $\omega$ is a complex number). This does not converge, but by subtracting off the "divergent part" $$\sum_{m^2+n^2\neq0}\frac{1}{(m+\omega n)^2}$$

you get a doubly-periodic meromorphic function. By taking $\omega\rightarrow\infty$, this becomes the singly-periodic function in your question.

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Actually, I started looking at this sum because I wondered what the real analog of the Weierstrass elliptic function should be. My above guess was motivated by the assumption that if the sum WASN'T a known function, it'd have a name, like the 'Weierstrass periodic function'. –  Greg Muller Jan 6 '10 at 23:56

As long as we're talking about the Weierstrass function, consider the parallels between the following:

1) Given a lattice $\Gamma$ in $\mathbb{R}$, the quotient $\mathbb{R}/\Gamma$ is topologically a circle. One way to describe it is to write down some functions on $\mathbb{R}$ which are invariant under $\Gamma$ and consider their image. It's natural to do this by writing down some arbitrary function on $\mathbb{R}$ and then averaging over $\Gamma$, and choosing $\frac{1}{x^2}$ gives a function which is closely related to $\sin \theta$. Now the map $\mathbb{R}/\Gamma \to (\sin' \theta, \sin \theta)$ has image the real variety $x^2 + y^2 = 1$.

2) Given a lattice $\Gamma$ in $\mathbb{C}$, the quotient $\mathbb{C}/\Gamma$ is topologically a torus. One way to describe it is to write down some functions on $\mathbb{C}$ which are invariant under $\Gamma$ and consider their image. Again it's natural to do this by writing down some arbitrary function on $\mathbb{C}$ and then averaging over $\Gamma$, and choosing $\frac{1}{z^2}$ gives a function which is closely related to the Weierstrass function $\wp(z)$. Now the map $\mathbb{C}/\Gamma \to (\wp'(z), \wp(z))$ has image the complex variety $y^2 = 4x^3 - g_2 x - g_3$. (Of course, here it is much more natural to expect that this strategy works because we know that compact Riemann surfaces are algebraic.)

Indeed, it's possible to view circles as "degenerate elliptic curves" in the following precise sense: every elliptic curve over $\mathbb{C}$ can be put in the form

$$x^2 + y^2 = 1 + a^4 x^2 y^2$$

for some $a \neq 0 \in \mathbb{C}$. The elliptic functions which parameterize these varieties are related to the Jacobi elliptic functions, and the above form is (essentially) Edwards normal form. As Edwards notes, one of the many conceptual advantages of this normal form is that as $a \to 0$, the group law degenerates to the angle addition formula for the sine and cosine! There is a paper by Franz Lemmermeyer which explores this degeneration from an arithmetic perspective.

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Of course, circles are also degenerate elliptic curves in the sense that a torus whose annular radius (?) shrinks to zero becomes a circle, but I think it's nice to demonstrate the degeneracy algebraically because, among other things, it's hidden by Weierstrass form. –  Qiaochu Yuan Feb 23 '10 at 4:38

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