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Context: I am currently reading Watson's 1961 paper "Goodness-Of-Fit Tests on a Circle". At a certain step, He produces the following relation (Equation (18) to (20) if you are able to read the actual paper):

$$\frac{\sqrt{\theta/2}}{\sin(\sqrt{\theta/2})}=2\sum_{m=1}^{+\infty}\frac{(-1)^{m-1}}{1-\frac{\theta}{2m^2\pi^2}}$$

(by expanding Euler's infinite product formula for $\sin$).

Question: the general term of the above series representation doesn't converge to $0$. What sense can we give to this sum ?

Additional context: He then apply the inverse Laplace transform term by term, so the notion of convergence should be "strong enough" to allow some kind of linearity.

What I think, so far:

  • I displayed the graph of both functions : enter image description here The red graph is the left hand side of the equality, extended for negatives values of $\theta$ by $$\frac{\sqrt{-\theta/2}}{\sinh(\sqrt{-\theta/2})}$$ (dashed line), and the green graph is the right hand side, truncated at the first twenty terms (even pushing a bit more doesn't give a perfect fit).
  • I know Watson is well known for computing asymptotic series expansion (in the sense introduced by Poincaré) of Laplace transforms, could it be such expansion (for small $\theta$)?
  • Is he using some alternative notion of convergence (Borel summation for example, which is often used in the context of Laplace transforms)?
Kolakoski54
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  • The identity after differentiating the product of sine is $\pi/\sin(\pi z) =\frac1z+\sum_{n=1}^\infty\frac{2z(-1)^n}{z^2-n^2}$. I'm confused how did Watson get the form in your question –  Mar 09 '25 at 10:41
  • I think he expanded this formula for $\sin$: https://math.stackexchange.com/questions/674769/sinx-infinite-product-formula-how-did-euler-prove-it – Kolakoski54 Mar 09 '25 at 10:46
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    I don’t have access to the paper from Watson, but if you differentiate Euler’s product form of sine logarithmically you will get the identity from my previous comment –  Mar 09 '25 at 13:02

1 Answers1

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I think I know what Watson meant.

Take the product of sine by Euler $$\sin(\pi x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$

Taking log and then differentiating we get the Mittag-Leffler pole expansion of cot $$\pi \cot(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}$$ Which by elementary trigonometric identities we also have one for csc $$\pi \csc(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z(-1)^n}{z^2-n^2}\tag{1}$$

Now we take the equation by Watson from your question. Take $\sqrt{\frac{\theta}{2}}=\pi z$, so we recast the equation as

$$\frac{\pi z}{\sin(\pi z)}=2\sum_{m=1}^{\infty}(-1)^{m-1}\left(1+\frac{z^2}{m^2-z^2}\right)$$ Which is the identity $(1)$ if we agree to assign the divergent sum $\sum_{n=1}^{\infty}(-1)^{n-1}$ the value $\frac{1}{2}$. Here are a few interpretations for this result

$1.$ If we take the average of the partial sums $\sum (-1)^{n-1}$ we get $\frac{1}{2}$. (See Cesàro summation)

$2.$ It is the analytic continuation of the geometric series $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$

$3.$ It is the analytic continuation of the Dirichlet eta function at $\eta(s)$ at $s=0$. Which the limit $\lim_{s\to 0}\eta(s)$ exists rigorously

  • Thank you for your answer! I'm interested in how he managed to apply the inverse Laplace transform: I don't think we can (in general) apply such linear operator term by term for "non-conventional" type of convergence... – Kolakoski54 Mar 09 '25 at 17:37