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I have accrossed this sum which is defined as :$\sum_{-\infty}^{+\infty}\frac{\exp(-n^2)}{1-4n^2}$ , Wolfram alpha assume that series is converges and gives $\approx 0.75229789\cdots$ , really I have tried to present that value in closed form using Inverse symbolic calculator but i didn't succeeded , Now my question is : to give the value of the titled series in the closed form ?

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    $\begingroup$ Do you have any reason to think that this has a decent closed form? What do you mean by "integral representation"? $\endgroup$ – robjohn Jun 26 '18 at 19:55
  • $\begingroup$ Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. $\endgroup$ – robjohn Jun 26 '18 at 19:55
  • $\begingroup$ Most likely this will be related to the integral/derivatives of theta functions: mathworld.wolfram.com/JacobiThetaFunctions.html (if that counts as a closed form) $\endgroup$ – Alex R. Jun 26 '18 at 19:57
  • $\begingroup$ @robjohn: there is a nice closed form. This is a problem recently proposed by Cornel Ioan Valean. $\endgroup$ – Jack D'Aurizio Jun 27 '18 at 11:23
  • $\begingroup$ @JackD'Aurizio: That is nice. That should have been included in the question for context. $\endgroup$ – robjohn Jun 27 '18 at 13:14
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EDIT 29.06.18 13:15

Correction

The previously provided closed expression $s_{c}$ was shown to be incorrect by some commenters.

As Mariusz pointed out correctly there are jumps in the function $p(z)$ (see (d8)) at $z=k \;\pi$

Correcting for $m$ of those jumps the closed expression is given by the (extremely fast converging) series

$$s_{c}(m) = s_{c}+\frac{\pi }{2 \sqrt[4]{e}} \sum _{k=1}^m (-1)^k \left(\text{erfi}\left(\frac{1}{2}+i \pi k\right)+\text{erfi}\left(\frac{1}{2}-i \pi k\right)\right)$$

Numerical examples

Letting $sN(100) = s$ up to 100 valid digits we obtain for the differences $d(m) = s_{c}(m) - sN(100)$ for $m=0..3$ the following

$$ \left\{-\frac{3.996}{10^6},-\frac{1.54}{10^{19}},-\frac{2.5929}{10^{41}},-\frac{1.456}{10^{71}}\right\}$$

Summarizing: instead of a true closed expression for $s$ we have effectively transformed one sum into another sum, which, however converges very fast.

The original question is now to be repeated for the latter sum.

Remark

Independently of this correction, the discovered strange numerical behaviour Mathematica deserves further study.

Note added 1. July: Some steps have already been taken here https://mathematica.stackexchange.com/questions/176240/bug-in-analytical-expression-of-integral-containing-abs-function/176342#176342

EDIT 28.06.18 14:00

Doubts have been raised in comments that my result might be wrong. The arguments given seem to rely on the numerical evaluation in Mathematica. Therefore I have extended the derivation to show more steps so that possible flaws can be detected.

Post as of 27.06.18

I have found the closed expression for the sum

$$s=\sum _{n=-\infty }^{\infty } \frac{\exp \left(-n^2\right)}{1-4 n^2}$$

It is given by

$$s_{c}=\frac{\pi\; \text{erfi}\left(\frac{1}{2}\right)}{2 \sqrt[4]{e}} $$

Derivation see below.

Numerically, the first 60 digits of $s_{c}$ according to Mathematica 10.1 are

$$N(s_{c}) = 0.75229|3902402569849043685417920199342618157039554017947019766$$

while, as has been pointed out in two comments, $s_{c}$ differs from $s$ numerically already in the fifth decimal digit:

$$N(s) = 0.75229|7898472243144830594123416216568518483359108518774883675$$

Something is wrong here.

Original post

As a partly solution we give here an integral representation of the sum in question

$$s=\sum _{n=-\infty }^{\infty } \frac{\exp \left(-n^2\right)}{1-4 n^2}$$

Splitting the sum into the term with $n=0$ and observing the symmetry of the summands $s$ can be witten as

$$s = 1-2 p\tag{1}$$

where

$$p = \sum _{n=1}^{\infty } \frac{\exp \left(-n^2\right)}{4 n^2-1}\tag{2}$$

Writing the denomintor for $n \ge 1$ as an integral

$$\frac{1}{4 n^2-1} = \int_0^{\infty } \exp \left(-t \left(4 n^2-1\right) \right) \, dt \tag{3}$$

leads under the integral to the sum

$$ \sum _{n=1}^{\infty } \exp \left(-\left(4 n^2-1\right) t-n^2\right) $$

Which can be evaluated in terms of the Jacobi Theta function:

$$\frac{1}{2} e^t \left(\vartheta _3\left(0,e^{-4 t-1}\right)-1\right)\tag{4}$$

Using (3) we find the integral representation

$$p=\int_0^{\infty } \frac{1}{2} e^t \left(\vartheta _3\left(0,e^{-4 t-1}\right)-1\right) \, dt\tag{5}$$

Derivation of the closed expression

The ingredients are easy to verify:

Partial fraction decomposition gives instead of (3):

$$\frac{1}{4 n^2-1} = \int_0^{\infty } \sinh (t) \exp (-2 n t) \, dt\tag{d1}$$

Under the Fourier transform the exponent becomes linear in $n$

$$e^{-n^2} = \frac{1}{\sqrt{\pi }} \int_{-\infty }^{\infty } \exp \left(-x^2\right) \exp (2 i n x) \, dx\tag{d2}$$

Under the two integrals the sum for $p$ is linear in the Exponent, i.e. it is a geometric sum, and hence can easily be done:

$$\frac{1}{\sqrt{\pi }}\sum _{n=1}^{\infty } \exp \left(-x^2\right) \sinh (t) \exp (-2 n t) \exp (2 i n x) = \frac{e^{-x^2+2 i x} \sinh (t)}{\sqrt{\pi } \left(e^{2 t}-e^{2 i x}\right)}\tag{d3}$$

Now the t-integral is solved by Matematica assuming that $x$ is real to give

$$\int_0^{\infty } \frac{e^{-x^2+2 i x} \sinh (t)}{\sqrt{\pi } \left(e^{2 t}-e^{2 i x}\right)} \, dt = \frac{e^{-x^2} \left(1+2 i \sin (x) \tanh ^{-1}\left(e^{i x}\right)\right)}{2 \sqrt{\pi }}\tag{d4}$$

Last but not least, the x-integral between $-\infty$ and $+\infty$ can be written, using the symmetry of the integrand as

$$p=\int_0^{\infty } \frac{e^{-x^2} \left(1-i \sin (x) \left(\tanh ^{-1}\left(e^{-i x}\right)-\tanh ^{-1}\left(e^{i x}\right)\right)\right)}{\sqrt{\pi }} \, dx\tag{d5}$$

Observing that

$$i \left(\tanh ^{-1}\left(e^{-i x}\right)-\tanh ^{-1}\left(e^{i x}\right)\right)=\frac{\pi }{2} \; \text{sgn}(\sin (x))\tag{d6}$$

the x-integral becomes

$$p=\frac{1}{2 \sqrt{\pi }}\int_{-\infty }^{\infty } e^{-x^2} \left(1-\frac{1}{2} \pi \left| \sin (x)\right| \right) \, dx\tag{d7}$$

where we have returned to the symmetric form, correcting this by the factor $\frac{1}{2}$ in front

Mathematica refused to do this integral directly but it was successful with the finite integral, with some $z\gt 0$

$$p(z)=\frac{1}{2 \sqrt{\pi }}\int_{-z }^{z} e^{-x^2} \left(1-\frac{1}{2} \pi \left| \sin (x)\right| \right) \, dx$$

$$= \frac{1}{8} \left(4 \text{erf}(z)+\frac{\pi \left(-2 \text{erfi}\left(\frac{1}{2}\right)+\left(\text{erfi}\left(\frac{1}{2}+i z\right)+\text{erfi}\left(\frac{1}{2}-i z\right)\right) \csc (z) \left| \sin (z)\right| \right)}{\sqrt[4]{e}}\right)\tag{d8}$$

Now the Limit $z\to\infty$ has to be taken. Again Mathematica refused but it did this asymtotic series expansion about $z=\infty$

$$\frac{i \sqrt{\pi } e^{-z^2-i z} \csc (z) \left| \sin (z)\right| }{8 z}-\frac{i \sqrt{\pi } e^{-z^2+i z} \csc (z) \left| \sin (z)\right| }{8 z}-\frac{\pi \text{erfi}\left(\frac{1}{2}\right)}{4 \sqrt[4]{e}}-\frac{e^{-z^2}}{2 \sqrt{\pi } z}+\frac{1}{2}\tag{d9}$$

Letting now $z\to\infty$ gives finally

$$p = \frac{1}{2}-\frac{\pi \text{erfi}\left(\frac{1}{2}\right)}{4 \sqrt[4]{e}}\tag{d10}$$

and the intial sum becomes

$$s = 1 - 2 p = \frac{\pi \text{erfi}\left(\frac{1}{2}\right)}{2 \sqrt[4]{e}}$$

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  • $\begingroup$ nice and interesting answer (+1)! $\endgroup$ – G Cab Jun 26 '18 at 21:02
  • $\begingroup$ it was already very interesting in the first version, now it is super; but pls. elucidate better the steps leading to the new integral and to erfi $\endgroup$ – G Cab Jun 28 '18 at 0:27
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    $\begingroup$ Here is the source of the error in mathematica. Define a = 1 - 2 NIntegrate[1/(2 Sqrt[[Pi]]) Exp[-x^2] (1 - 1/2 [Pi] Abs[Sin[x]]), {x, -zmax, zmax}]; and b = 1 - 2 Integrate[1/(2 Sqrt[[Pi]]) Exp[-x^2] (1 - 1/2 [Pi] Abs[Sin[x]]), {x, -z, z}]; By definition we should a=b when z=zmax, but N[a - b /. z -> zmax] evaluates to $\sim 4\cdot 10^{-6}$ for zmax $= 50$ (the difference between your result and the true sum. $\endgroup$ – Winther Jun 28 '18 at 14:08
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    $\begingroup$ The difference between these two expressions happens only for $z >\pi$. Thus it's probably a condition in the analytical expression that it only holds for $z<\pi$ that mathematica fails to list. (In the code above compare zmax=3+1/10+4/100 to zmax=3+1/10+5/100). If I were to guess it's likely due to the Abs[] function. $\endgroup$ – Winther Jun 28 '18 at 14:21
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    $\begingroup$ @ zeraoulia rafik Ths is the second occasion here that Cornel Iona valeen is mentioned as the "authority in the background". Could you please ask him to reveal his knowledge? I would greatly appreciate learning more. $\endgroup$ – Dr. Wolfgang Hintze Jun 30 '18 at 8:23
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I found general formula. It's not closed form solution only approximation for the sum:

$$\sum _{j=-\infty }^{\infty } \frac{\exp \left(-j^2\right)}{1-4 j^2}\approx \frac{\pi \sum _{k=1}^n \left(\frac{\text{erfi}\left(\frac{1}{2}\right)}{n}+(-1)^k \text{erfi}\left(\frac{1}{2}-k i \pi \right)+(-1)^k \text{erfi}\left(\frac{1}{2}+k i \pi \right)\right)}{2 \sqrt[4]{e}}$$

for $n=1$ it's only correct for 18 digits.

for $n=2$ it's only correct for 40 digits.

for $n=3$ it's only correct for 70 digits.

for $n=4$ it's only correct for 109 digits.

if $n>4$ then it's better approximation and more correct digits.

Derivation of the formula

Borrowing integral $(d7)$ form user Dr. Wolfgang Hintze and integrating with Mathematica help:

$$\int \frac{\exp \left(-x^2\right) \left(1-\frac{1}{2} \pi \left| \sin (x)\right| \right)}{2 \sqrt{\pi }} \, dx=\\\frac{4 \sqrt[4]{e} \text{erf}(x)-\pi \left(\text{erfi}\left(\frac{1}{2}-i x\right)+\text{erfi}\left(\frac{1}{2}+i x\right)\right)+2 \pi \left(\text{erfi}\left(\frac{1}{2}-i x\right)+\text{erfi}\left(\frac{1}{2}+i x\right)\right) \theta (\sin (x))}{16 \sqrt[4]{e}}+C$$

taking limit in jumps points:

$$\left(\underset{x\to \pi ^-}{\text{lim}}\text{int}-\underset{x\to 0^+}{\text{lim}}\text{int}\right)+\left(\underset{x\to \infty }{\text{lim}}\text{int}-\underset{x\to (2 \pi )^+}{\text{lim}}\text{int}\right)+\left(\underset{x\to (2 \pi )^-}{\text{lim}}\text{int}-\underset{x\to \pi ^+}{\text{lim}}\text{int}\right)+\left(\underset{x\to (-2 \pi )^-}{\text{lim}}\text{int}-\underset{x\to -\infty }{\text{lim}}\text{int}\right)+\left(\underset{x\to 0^-}{\text{lim}}\text{int}-\underset{x\to (-\pi )^+}{\text{lim}}\text{int}\right)+\left(\underset{x\to (-\pi )^-}{\text{lim}}\text{int}-\underset{x\to (-2 \pi )^+}{\text{lim}}\text{int}\right)=\\\frac{1}{2}-\frac{\pi \left(\text{erfi}\left(\frac{1}{2}\right)-\text{erfi}\left(\frac{1}{2}-i \pi \right)-\text{erfi}\left(\frac{1}{2}+i \pi \right)+\text{erfi}\left(\frac{1}{2}-2 i \pi \right)+\text{erfi}\left(\frac{1}{2}+2 i \pi \right)\right)}{4 \sqrt[4]{e}}$$

based on this, it was possible to deduce the formula.

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  • $\begingroup$ @ Mariusz Iwaniuk I saw your contribution only a few minutes ago. Sorry, was so busy creating my corrected solution. So we have arived independently at the same result. Still in my opinion, Mathematica is buggy with the finite integral, d7: it generates spurious jumps, without warning. Integrating a continuous function must not lead to jumps. No, it is the ancient story that a finite integral can be wrong if the antiderivative has jumps ... But just returning a value for a finite integral without warning even if the antiderivative has jumps can be heavily miseading. What do you think? $\endgroup$ – Dr. Wolfgang Hintze Jun 29 '18 at 15:13
  • $\begingroup$ @Dr.WolfgangHintze. You may be interested:mathematica.stackexchange.com/questions/176240/… People make mistakes, it is natural.Yes MMA is buggy I'm sure of it. Over 2 years I sent them to support over 250 bugs. Maple is not better either.About:antiderivative,jumps,well, it is probably difficult to implement for different cases. We only hope that there will be as few bugs as possible. $\endgroup$ – Mariusz Iwaniuk Jun 29 '18 at 15:42
  • $\begingroup$ @ Mariusz Iwaniuk Thanks for the link which the author himself didn't bother to tell us here. He takes the problem from my (painful) solution without mentioning the source :-( And we still don't know the answer to the original question ... $\endgroup$ – Dr. Wolfgang Hintze Jun 29 '18 at 18:04

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