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 ?

  • 2
    $\begingroup$ Do you have any reason to think that this has a decent closed form? What do you mean by "integral representation"? $\endgroup$
    – robjohn
    Commented Jun 26, 2018 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
    Commented Jun 26, 2018 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.
    Commented Jun 26, 2018 at 19:57
  • $\begingroup$ @robjohn: there is a nice closed form. This is a problem recently proposed by Cornel Ioan Valean. $\endgroup$ Commented Jun 27, 2018 at 11:23
  • $\begingroup$ @JackD'Aurizio: That is nice. That should have been included in the question for context. $\endgroup$
    – robjohn
    Commented Jun 27, 2018 at 13:14

2 Answers 2


EDIT 02.05.21

Inspired by a recent comment of @stocha I have found a simpler derivation of the final result, without having to consider jumps in the antiderivative.

From $(d7)$ and $(1)$ we have

$$s := \sum_{n=-\infty}^{\infty}\frac{e^{-n^2}}{1-4n^2}= \sqrt{\pi} \int_{0}^{\infty} e^{-x^2} | \sin(x)|\,dx\tag{e1}$$

Now we break up the integration range into intervals from $ \pi k$ to $\pi(k+1)$, $k=0,1,2,...$. This serves to remove the absolute value function from the $\sin$ and we get

$$\begin{align}\frac{s}{\sqrt{\pi}} &= ​\int_{0}^{\pi} e^{-x^2} \sin(x)\,dx - \int_{\pi}^{2\pi} e^{-x^2} \sin(x)\,dx\\ +& \int_{2\pi}^{3\pi} e^{-x^2} \sin(x)\,dx - \int_{3\pi}^{4\pi} e^{-x^2} \sin(x)\,dx + ...\\ =& \sum_{k\ge 0} (-1)^k \int_{\pi k}^{\pi(k+1)} e^{-x^2} \sin(x)\,dx\end{align}\tag{e2}$$


$$\int e^{-x^2} \sin(x)\,dx=\frac{\sqrt{\pi}}{4 e^{\frac{1}{4}}}\left(\text{erfi}(\frac{1}{2}+ i x)+ \text{erfi}(\frac{1}{2}- i x) \right)\tag{e3}$$

where $\text{erfi}(z)= - i \text{erf}(i z)$ and $\text{erf}$ is the error function so that the one sided integral in $(e1)$ gives rise to this double sided alternating sum

$$s = \frac{\pi}{2 e^{\frac{1}{4}}}\sum_{k=-\infty}^{\infty}(-1)^k \text{erfi}(\frac{1}{2} + i \pi k)\tag{e4}$$

It can also be written as

$$s = \frac{\pi}{2 e^{\frac{1}{4}}} \text{erfi}(\frac{1}{2} )+\frac{\pi}{e^{\frac{1}{4}}}\sum_{k=1}^{\infty}(-1)^k \Re\left( \text{erfi}(\frac{1}{2} + i \pi k)\right)\tag{e5}$$

The asymptotic behaviour of the summand for $k\to \infty$ is given by

$$\frac{\pi}{2 e^{\frac{1}{4}}}\Re\left( \text{erfi}(\frac{1}{2} + i \pi k)\right)\sim \frac{e^{-\pi ^2 k^2} \sin (\pi k)}{\pi ^{1/2} k}+O(\frac{1}{k^2})\tag{e6}$$

this shows the excellent convergence of the sum.

Using Cauchy's theorem. Useful or not?

In a (desparate) attempt to find a closed expression we could write both two-sided sums $(e1)$ and $(e4)$ in the form a contour integral with the kernel $\pi \cot(\pi n)$ and $\frac{\pi}{ \sin(\pi k)}$, repectively, but I see no use in it as both integrands diverge for large imaginary argument. Maybe someone else can pusue this approach?

EDIT 29.06.18 13:15


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.


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

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


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

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

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

  • $\begingroup$ nice and interesting answer (+1)! $\endgroup$
    – G Cab
    Commented Jun 26, 2018 at 21:02
  • 1
    $\begingroup$ 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
    Commented Jun 28, 2018 at 14:08
  • 1
    $\begingroup$ 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
    Commented Jun 28, 2018 at 14:21
  • 2
    $\begingroup$ @ 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. $\endgroup$ Commented Jun 30, 2018 at 8:23
  • 1
    $\begingroup$ @ stocha Very good. And thanks for inspiring me some days ago to find a simple derivation. See my recent EDIT. $\endgroup$ Commented May 2, 2021 at 9:25

I found general formula. It's not closed form solution only approximation for the sum:

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

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.

  • $\begingroup$ @ 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? $\endgroup$ Commented Jun 29, 2018 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. $\endgroup$ Commented Jun 29, 2018 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 ... $\endgroup$ Commented Jun 29, 2018 at 18:04
  • $\begingroup$ @ Mariusz Iwaniuk There seems to be a typo in your main formula. Shouldn't the denominator of $erfi(\frac{1}{2})$ be 1 rather than $n$? $\endgroup$ Commented May 2, 2021 at 8:47
  • $\begingroup$ @Dr.WolfgangHintze.Yes you are right.I edit my answer. $\endgroup$ Commented May 2, 2021 at 9:31

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