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This question will be very similar to:

On $$\mathrm{\sum\limits_{x=1}^\infty Ci(x)}$$

and

On $$\mathrm{\sum_{x\in\Bbb Z}sech(x), \sum_{x=1}^\infty csch(x)}$$

all of which had closed forms, but it will use this Airy Ai function definition which is a type of Bessel function.

Note that I will actually focus on 2 constants which each converge slowly. I will use $\sum\limits_{n=a}^b A_n=\sum\limits_a^b A_n$ for shorthand and use the following notations. The reason these are split up is because one of the constants may possibly diverge. Note that $\mathrm A_{0,1}$ both have a lower bound of $0$, so the x=$0$ term will be subtracted:

$$\mathrm{A_0=\sum_{-\infty}^0 Ai(x), A_1=\sum_0^\infty Ai(x),A=A_0+A_1-Ai(0)=\sum_{-\infty}^\infty Ai(x)}$$

Here is a graph of the summand: enter image description here

Here is a possible Abel-Plana formula computation.

$$\mathrm{A_0=\sum_0^{\infty}Ai(-x)\mathop=^{Abel}_{Plana}\frac12 Ai(-0)+\int_0^\infty Ai(-x) \,dx+\int_0^\infty\frac{Ai(- -ix)-Ai(- ix)}{e^{2\pi x}-1}\,dx=\frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac23+ \int_0^\infty\frac{Ai(ix)-Ai(- ix)}{e^{2\pi x}-1}\,dx}$$

$$\mathrm{A_1=\sum_0^\infty Ai(x)= \mathop=^{Abel}_{Plana}\frac12 Ai(0)+\int_0^\infty Ai(x) \,dx+\int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx= \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac13+ \int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx}$$

This means that the conjectured answer is:

$$\mathrm{A=\sum_{x\in\Bbb Z}Ai(x)=A_0+A_1-Ai(0) \mathop=^{Abel}_{Plana} \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac23+ \int_0^\infty\frac{Ai(ix)-Ai(- ix)}{e^{2\pi x}-1} \, dx + \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac13+ \int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx-Ai(0)= \frac{1}{3^\frac23 Γ\left(\frac23\right)}+ 1+\int_0^\infty\frac{Ai(ix)-Ai( -ix)+Ai(-ix)-Ai(-ix)}{e^{2\pi x}-1}\,dx-\frac{1}{3^\frac23 Γ\left(\frac23\right)}}=1 $$

My final conjecture is the following with alternate forms. Note that some simplifications are possible, but change the definition. The following also uses Hypergeometric functions:

$$\mathrm{1\mathop=^?A=\sum_{-\infty}^\infty Ai(x)=\sum_{x\in\Bbb Z}Ai(x)= \sum_{-\infty}^\infty \left(\frac{\,_0F_1\left(\frac23,\frac{x^3}{9}\right)}{3^\frac23Γ\left(\frac23\right)}-\frac{\,_0F_1\left(\frac43,\frac{x^3}{9}\right)}{\sqrt[3]3 Γ\left(\frac13\right)}\right)} =\sum_{-\infty}^\infty\left(\left(x^\frac32\right)^\frac13I_{-\frac13}\left(\frac{2x^\frac32}{3}\right)-x \left(x^\frac32\right)^{-\frac13} I_\frac13\left(\frac{2x^\frac32}{3}\right)\right) $$

There are also variations with the Airy Bi function, and the Scorer functions Gi and Hi. Please correct me and give me feedback!

Тyma Gaidash
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  • Don't you get $1$ when you add up the two Abel-Plana results and subtract the extra $\operatorname{Ai}(0)$? The integrals just cancel each other out. – Gary Sep 17 '21 at 00:34
  • @Gary This formula implies that the sum over the integers is just the area under the function, so the sum is the same as the integral. Good observation. I used it in the problem. The formula does not always work. – Тyma Gaidash Sep 17 '21 at 00:56
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    $\operatorname{Ai}(\pm ix)$ grows too fast for the Abel-Plana to be applicable. – metamorphy Sep 17 '21 at 04:03
  • I don't even see a clearway justification that $\sum_{n=-\infty}^{-1}\operatorname{Ai}(n)$ converges. According to the asymptotics, this is equivalent to the convergence of $\sum_{n=1}^\infty\frac1{n^{1/4}}\sin\left(\frac23 n^{3/2}+\frac\pi4\right)$, which is not at all obvious. (And I don't trust your numerical experiments not related to regularised things.) – metamorphy Sep 22 '21 at 15:37
  • Just for fun, this is the partial sum plot of your sine series. – Тyma Gaidash Sep 22 '21 at 17:31
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    Linking to this question (the convergence issue is still open). – metamorphy Jul 27 '22 at 03:59
  • @metamorphy Using $\sum_{n=1}^\infty \frac{\sin(n^a)}{n^b}$ converging if $b>\max(a,1-a),a>0$, the sum is $\frac1{\sqrt 2}\sum_{n=1}^\infty \frac{\sin(\frac23n^{3/2})}{n^\frac14}+\frac1{\sqrt2} \sum_{n=1}^\infty \frac{\cos(\frac23n^{3/2})}{n^\frac14}$. However, $\sum_{n=1}^\infty \frac{\sin(m n^a)}{n^b}$ and $\sum_{n=1}^\infty \frac{\cos(m n^a)}{n^b}$’s convergence is unknown – Тyma Gaidash Jul 27 '22 at 13:16

2 Answers2

2

I got something in terms of a theta function: \begin{align*} \sum\limits_{n = - \infty }^\infty \operatorname{Ai}(n) & = \sum\limits_{n = - \infty }^\infty {\frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\left( {\tfrac{1}{3}t^3 + nt} \right)} \right)dt} } \\ & = \frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\tfrac{1}{3}t^3 } \right)\sum\limits_{n = - \infty }^\infty {\exp \left( {i n t} \right)} dt} \\ & = \frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\tfrac{1}{3}t^3 } \right)\theta _3 \!\left( {\tfrac{t}{2},1} \right)dt} \\ & = \frac{1}{\pi }\int_0^{ + \infty } {\cos \left( {\tfrac{1}{3}t^3 } \right)\theta _3 \!\left( {\tfrac{t}{2},1} \right)dt} . \end{align*}

Gary
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    Thanks. Also, the function is technically not defined for:$\vartheta_3\left(\frac x2,1\right)$ as $\vartheta_3\left(\frac x2,z\right),|z|<1$ as seen here. Using z=.99, the integral evaluates to: $3.14157$, but we have a factor of $\frac1\pi$, so this means the Abel Plana formula should work and the $\sum_{-\infty}^{\infty} Ai(x)=1$ while the unregularized sum is $\pi$. Right? – Тyma Gaidash Sep 16 '21 at 15:57
  • one last thing can you help me with:$$\sum_{x=-\infty}^0 \text{Bi}(x)$$? Using the other main Airy function? – Тyma Gaidash Sep 16 '21 at 16:27
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    You can use http://dlmf.nist.gov/9.5.E3 to obtain an expression in terms of two integrals. – Gary Sep 16 '21 at 16:39
  • A few words on why I don't think this should be accepted. In the distributional sense, $$\sum_{n=-\infty}^\infty\exp(int)=2\pi\sum_{n=-\infty}^\infty\delta(t-2n\pi)$$ is known as Dirac comb, but this is not applicable here (the integration is outside of an application of the theory of distributions, and the resulting series doesn't converge). I wonder if this can be "fixed". – metamorphy Oct 06 '21 at 20:08
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This will be a section for the “variations” mentioned in the question:

Corollary for Abel-Plana formula:

Let $f(x)$ fit the conditions for the Abel-Plana, $f(ix)-f(-ix)+f(-ix)-f(ix)=0$, and $f(0)$ exist then:

$$\sum_{\Bbb Z} f(x)=-f(0)+\sum_0^{\infty} f(-x)+\sum_0^\infty f(x)=-f(0)+\frac12 f(-0)+\int_0^\infty f(-x)dx+i\int_0^\infty \frac{f(- - ix)-f(-ix)}{e^{2\pi x}-1}dx+\frac12 f(0)+\int_0^\infty f(x) dx+i\int\frac{f(-ix)-f(ix)}{e^{2\pi x}-1}dx=-f(0)+f(0)+\int_0^\infty f(-x)+f(x) dx+i \int_0^\infty \frac{f(ix)-f(-ix)+f(-ix)-f(ix)}{e^{2\pi x}-1}dx=\int_0^\infty f(x)+f(-x) dx= \int_{\Bbb R} f(x)dx$$

Therefore a nice, not so useless, final result is:

$$\sum_{\Bbb Z} f(x)=\int\limits_{\Bbb R} f(x) dx$$ Ai variation:

Using @Gary’s result and the numerical integration shows that $A=1$. Now let us define the uregularized main integral definitions of the Airy Ai function. Note that I use $\Bbb R=[-\infty,\infty]$ and the derived corollary:

$$\sum_{x\in\Bbb Z} \int_0^\infty \cos\left(\frac{t^3}{3}+tx\right)dt= \int_{-\infty}^\infty \int_0^\infty \cos\left(\frac{t^3}{3}+tx\right)dt\,dx=\pi, \sum_{x\in\Bbb Z} \int_{\Bbb R} e^{i\left(\frac{t^3}{3}+tx\right)}dt= \iint_\limits{\Bbb R^2} e^{i\left(\frac{t^3}{3}+tx\right)}dt\,dx=2\pi$$

Bi version using @Gary’s suggestion of DLMF 9.5.3 and my finding of the Scorer function integral representations

$$\pi\sum_{-\infty}^0\text{Bi}(x)= \sum_{x=0}^\infty \int_0^\infty e^{-\frac{t^3}{3}-tx}dt+ \sum_{x=0}^\infty \int_0^\infty \sin\left(\frac{t^3}{3}+tx\right)dt= \pi \sum_{x=0}^\infty (\text{Gi}(x)+\text{Hi}(-x))$$

Тyma Gaidash
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