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I came across this problem while doing some simplifications.

So, I like to ask

QUESTION. Is there a closed formula for the evaluation of this series? $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}$$ where the sum runs over all pairs of positive integers that are relatively prime.

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    $\begingroup$ @user64494 why? "infinite sum" is a correct math term, and refers to the sum of a family of numbers, which is the present case. "Sum of a series" is a correct term, yet not applicable here, since the OP's one is not a series. $\endgroup$
    – Pietro Majer
    Commented Aug 8, 2021 at 17:53
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    $\begingroup$ @user64494 : As Pietro Majer said, "infinite sum" is a correct term, and $\sum_{x\in X}f(x)$ can be defined for any (say) real-valued function $f$ on any set $X$. One of a number of mutually equivalent ways to do that is as follows: $\sum_{x\in X}f(x):=\int_X f\,d\nu$, where $\nu$ is the counting measure on (the $\sigma$-algebra of all subsets of) $X$, if the integral exists. $\endgroup$
    – Iosif Pinelis
    Commented Aug 8, 2021 at 19:07
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    $\begingroup$ @T. Amdeberhan never follow suggestions from anonymous sources :) $\endgroup$
    – Pietro Majer
    Commented Aug 8, 2021 at 23:20
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    $\begingroup$ This question on MathStackExchange asked a similar question: math.stackexchange.com/questions/452028/… The method of solution given there in the accepted answer can be easily applied to the OPs case, similar approach using Mobius transformation. $\endgroup$
    – Nemo
    Commented Aug 9, 2021 at 7:25
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    $\begingroup$ @user64494 On the other hand, I think we all agree that an "infinitite sum" is an incorrect math term :-) $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 9, 2021 at 8:50

2 Answers 2

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Apply Möbius summation, the formula for $\sum_{n>=1}\cos(2\pi n x)/n^2$ to obtain:

$$11/4-45\zeta(3)/\pi^3=1.00543...\;$$

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    $\begingroup$ GH from MO has given much detail in the answer below. Let me just expand on what Henri Cohen referred to as Möbius summation: $\sum_{(a,b)=1}\frac{\cos(a/b)}{a^2b^2}=\sum_{m,n}\sum_{d\vert(m,n)}\frac{\mu(d)}{m^2n^2}=\sum_k\frac{\mu(k)}{k^4}\sum_{x,y}\frac1{x^2y^2}=\frac1{\zeta(4)}\sum_{x,y}\frac1{x^2y^2}$. $\endgroup$
    – T. Amdeberhan
    Commented Aug 9, 2021 at 13:22
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    $\begingroup$ Thank you for a cute answer, Henri Cohen. $\endgroup$
    – T. Amdeberhan
    Commented Aug 9, 2021 at 13:24
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Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the right hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$

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    $\begingroup$ Could you give a reference (or derivation) for the identity in the second display? $\endgroup$
    – Joshua Stucky
    Commented Aug 8, 2021 at 21:44
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    $\begingroup$ And a reference for the first identity, please. $\endgroup$
    – Iosif Pinelis
    Commented Aug 8, 2021 at 21:47
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    $\begingroup$ @JoshuaStucky: Reference added. $\endgroup$
    – GH from MO
    Commented Aug 8, 2021 at 21:59
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    $\begingroup$ @IosifPinelis: The first identity follows readily by grouping the pairs $(m,n)$ according to their greatest common divisor. That is, we write $m=ad$, $n=bd$ with $d=\gcd(m,n)$ and $\gcd(a,b)=1$. $\endgroup$
    – GH from MO
    Commented Aug 8, 2021 at 22:01
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    $\begingroup$ @GHfromMO: thank you for furnishing the details. $\endgroup$
    – T. Amdeberhan
    Commented Aug 9, 2021 at 13:23

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