Prove that $ \sum\limits_{n=-\infty}^\infty\frac{\cos\pi\sqrt{n^2+1}}{3+4n^2}=\int\limits_{-\infty}^\infty\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}dx $?

How one can prove that the infinite sum of this function equals its integral $$ \sum_{n=-\infty}^\infty\frac{\cos\pi\sqrt{n^2+1}}{3+4n^2}=\int_{-\infty}^\infty\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}dx\ ? \tag{1} $$

My analysis: Mathematica wasn't able to return any closed form for the integral or the sum. Then I checked this relation by numerical computations and it agreed to about 20 decimal places.

I know from this question Sum equals integral that the function $\text{sinc}\ x=\frac{\sin x}{x}$ has the same property $$ \int_{-\infty}^{+\infty} {\rm sinc}\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}\, n = \pi $$ I tried to find a closed form for the integral $(1)$ but couldn't.

Motivation: I was challenged by a friend to prove this relation. I'm curious how one can prove it?

Note: There has been a suggestion to straightforwardly apply Euler-MacLauren summation formula to prove this statement. Though I don't know why it can not be applied in this case, I checked numerically whether the sum equals the integral for the similar looking functions $f_1(x)=\frac{\cos\pi\sqrt{x^2+1}}{1+x^2}$ and $f_2(x)=\frac{\cos\pi\sqrt{x^2+1}}{2+x^2}$, but in both cases there was a difference of about 1% between the sum and the integral. In starck contrast to this, using the same algorithm for $\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}$ there wasn't any difference between the sum and the integral at least to 20 decimal places. So I think it is very unlikely that 1% error can be attributed to computational error.

Solution 1:

For any positive $a$, define

$$f_a(x) =\frac{\cos\pi\sqrt{x^2+1}}{x^2+a^2}$$

What you have observed is caused by the equality $$\sum_{n=-\infty}^\infty f_a(n) - \int_{-\infty}^\infty f_a(x) dx = \frac{2\pi}{a(e^{2\pi a} - 1)}\times \begin{cases} \cosh\pi\sqrt{a^2-1}, & a > 1\\ \\ \cos\pi\sqrt{1-a^2}, & a < 1 \end{cases} \tag{*1} $$ and the fact $$\cos\pi\sqrt{1-a^2} = \cos\frac{\pi}{2} = 0\quad\text{ when } a^2 = \frac34$$

To see why $(*1)$ is true, we use the fact $f_a(n)$ is an even function in $n$ to rewrite LHS of $(*1)$ as

$$2 \left[\sum_{n=0}^\infty f_a(n) - \left(\int_0^\infty f_a(x) dx + \frac12 f_a(0)\right)\right]$$

This is similar to what you will find in the Abel-Plana formula${}^{\color{blue}{[1]}}$,

For any function $f(z)$ which is

1. continuous on $\Re z \ge 0$ and analytic on $\Re z > 0$
2. $f(z) \sim o(e^{2\pi|\Im z|} )$ as $\Im z \to \pm \infty$, uniformly with respect to $\Re z$.
3. $f(z) \sim O(e^{2\pi|\Im z|}/|z|^{1+\epsilon})$ as $\Re z \to +\infty$ ${}^{\color{blue}{[2]}}$.

we have

$$\sum_{n=0}^\infty f(n) = \int_0^\infty f(x) dx + \frac12 f(0) + i \int_0^\infty \frac{f(it) - f(-it)}{e^{2\pi t}-1} dt\tag{*2}$$

However, $f_a(x)$ doesn't exactly satisfy the condition above. It has two poles at $\pm a i$. After a little bit of tweaking of the contour used in the proof of the Abel-Plana formula, one find:

$$\text{LHS}(*1) = 2i \lim_{\epsilon\to 0^{+}} \int_0^\infty \frac{f_a(it+\epsilon) - f_a(-it+\epsilon)}{e^{2\pi t} - 1} dt$$

For $t \ne a$, since $f_a(z)$ is even, the two pieces in $f_a(it+\epsilon) - f_a(-it+\epsilon)$ cancels out as $\epsilon \to 0^{+}$.
For $t \approx a$, the two pieces can be combined to a integral of $\frac{f(it)}{e^{2\pi t}-1}$ over a circle centered at $a$.
As a result, RHS reduces to

$$(2i)(2\pi i)\text{Res}_{t = a}\left[\frac{\cos\pi\sqrt{1-t^2}}{(a^2 - t^2)(e^{2\pi t} - 1)}\right] = \frac{2\pi}{a(e^{2\pi a}-1)}\times\begin{cases} \cosh\pi\sqrt{a^2-1}, & a > 1\\ \\ \cos\pi\sqrt{1-a^2}, & a < 1 \end{cases} $$ Back to the special case $a^2 = \frac{3}{4}$ which corresponds to the equality in question.

When $a^2 = \frac{3}{4}$, the "pole" of $f_a(z)$ at $z = \pm a i$ become removable singularities. The original version of Abel-Plana formula in $(*2)$ applies. Since $f_a(x)$ is even, last integral in $(*2)$ vanishes and the equality follows. This explain why the sum equal to the integral for $\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}$ but not other similar looking integrand like $\frac{\cos\pi\sqrt{x^2+1}}{1+x^2}$ or $\frac{\cos\pi\sqrt{x^2+1}}{2+x^2}$.

Notes

• $\color{blue}{[1]}$ For more details of Abel-Plana formula and its derivation, please refer to $\S 8.3$ of Frank W. J Olver's book: Asymptotics and Special Functions.
• $\color{blue}{[2]}$ In order to convert the AP formula on finite sum in Olver's book to infinite sum here, I have added a condition $(3)$ for this particular problem. The whole purpose of that is to force following limits to zero. $$\lim\limits_{b\to\infty} f(b) = 0\quad\text{ and }\quad\lim\limits_{b\to\infty}\int_0^\infty \frac{f(b+it)-f(b-it)}{e^{2\pi t} - 1}dt = 0$$ For other $f(z)$, if one can justifies these limits, we can forget condition $(3)$ and the AP formula remains valid.