Consider the equality: $$\sum _{h=0}^{\infty } \left(\sum _{k=1}^n x^{h n+k} f(\gcd (n,k))\right)=\frac{10 x^{10}+9 x^9+8 x^8+7 x^7+6 x^6+5 x^5+4 x^4+3 x^3+2 x^2+x}{x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1} \label{1}\tag{1}$$
where $$n=11$$
and $$f(n) = \sum_{d \mid n} \mu(d)d$$
Integrating the right hand side in Mathematica 14:
$$\int \frac{10 x^{10}+9 x^9+8 x^8+7 x^7+6 x^6+5 x^5+4 x^4+3 x^3+2 x^2+x}{x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1} \, dx$$
gives the root sum:
10 x - RootSum[
1 + #1 + #1^2 + #1^3 + #1^4 + #1^5 + #1^6 + #1^7 + #1^8 + #1^9 + \
#1^10 &, (
10 Log[x - #1] + 9 Log[x - #1] #1 + 8 Log[x - #1] #1^2 +
7 Log[x - #1] #1^3 + 6 Log[x - #1] #1^4 + 5 Log[x - #1] #1^5 +
4 Log[x - #1] #1^6 + 3 Log[x - #1] #1^7 + 2 Log[x - #1] #1^8 +
Log[x - #1] #1^9)/(
1 + 2 #1 + 3 #1^2 + 4 #1^3 + 5 #1^4 + 6 #1^5 + 7 #1^6 + 8 #1^7 +
9 #1^8 + 10 #1^9) &]
Integrating the left hand side instead:
$$\int x^{h n+k} \, dx=\frac{x^{h n+k+1}}{h n+k+1}$$
gives:
$$\sum _{h=0}^{\infty } \left(\sum _{k=1}^n \frac{x^{h n+k+1} f(\gcd (n,k))}{h n+k+1}\right) = \frac{-277200 x \, _2F_1\left(\frac{1}{11},1;\frac{12}{11};x^{11}\right)+2772 x^{10} \, _2F_1\left(\frac{10}{11},1;\frac{21}{11};x^{11}\right)+3080 x^9 \, _2F_1\left(\frac{9}{11},1;\frac{20}{11};x^{11}\right)+3465 x^8 \, _2F_1\left(\frac{8}{11},1;\frac{19}{11};x^{11}\right)+3960 x^7 \, _2F_1\left(\frac{7}{11},1;\frac{18}{11};x^{11}\right)+4620 x^6 \, _2F_1\left(\frac{6}{11},1;\frac{17}{11};x^{11}\right)+5544 x^5 \, _2F_1\left(\frac{5}{11},1;\frac{16}{11};x^{11}\right)+6930 x^4 \, _2F_1\left(\frac{4}{11},1;\frac{15}{11};x^{11}\right)+9240 x^3 \, _2F_1\left(\frac{3}{11},1;\frac{14}{11};x^{11}\right)+13860 x^2 \, _2F_1\left(\frac{2}{11},1;\frac{13}{11};x^{11}\right)-2520 \log \left(1-x^{11}\right)+277200 x}{27720} \label{2}\tag{2}$$
involving the hypergeometric function, which when $n$ is a prime number or when the left hand side of $\eqref{1}$ is of the form in the post about logarithms, appears to have the form:
$$\sum _{h=0}^{\infty } \left(\sum _{k=1}^n \frac{x^{h n+k+1} f(\gcd (n,k))}{h n+k+1}\right) = \sum _{k=2}^n \frac{x^k \, _2F_1\left(\frac{k}{n},1;\frac{k+n}{n};x^n\right)}{k}+(n-1) x \left(1-\, _2F_1\left(\frac{1}{n},1;\frac{n+1}{n};x^n\right)\right) \label{3}\tag{3}$$
The integral of the left hand side of $\eqref{1}$ which results in $\eqref{2}$ can be verified to be correct numerically:
Mathematica 14:
Clear[f, n, k, x, h, A, B]
nn = 11;
f[n_] := DivisorSum[n, MoebiusMu[#] # &]
n = nn;
S = Sum[Sum[f[GCD[n, k]]*x^(n*h + k), {k, 1, n}], {h, 0, Infinity}]
integral =
Sum[Sum[f[GCD[n, k]]*x^(1 + k + h n)/(1 + k + h n), {k, 1, n}], {h,
0, Infinity}];
x1 = 2;
x = x1;
a = integral;
x2 = 3;
x = x2;
b = integral;
N[b - a, 50]
NIntegrate[S, {x, x1, x2}, WorkingPrecision -> 50]
Output:
9.3079086593952989311454361373316908074201158341320 + 0.*10^-50 I
9.3079086593952989311454361373316908074201158341320
Question
Is this new? Or has it been used before? The goal is to express the reciprocal of the Riemann zeta function $$\frac{1}{\zeta(s)}$$ as a ratio of polynomials.
Maybe Möbius inversion of the integrals is all that is needed to make it general.
Any feedback or comments are welcome.
About the edit about the post about logarithms above:
Mathematica 14:
Clear[f, n, k, x, h, A, B]
nn = 11;
f[n_] := DivisorSum[n, MoebiusMu[#] # &]
n = nn;
S = Sum[Sum[If[k == n, 1 - n, 1]*x^(n*h + k), {k, 1, n}], {h, 0,
Infinity}]
Integrate[S, x]
integral =
Sum[Sum[If[k == n, 1 - n, 1]*x^(1 + k + h n)/(1 + k + h n), {k, 1,
n}], {h, 0, Infinity}];
FullSimplify[%]
Expand[(n - 1)*x*(1 - Hypergeometric2F1[1/n, 1, (n + 1)/n, x^n]) +
Sum[1/k*x^k*Hypergeometric2F1[k/n, 1, (n + k)/n, x^n], {k, 2, n}]]
FullSimplify[%% - %]
where FullSimplify[%% - %] returns 0 which means that the hypergeometric function expression of the integral in $\eqref{3}$ should be correct.
This is the Mathematica 14 code which appears to be correct when $n$ is the product of exactly $2$ prime numbers:
Clear[f, n, k, x, h, A, B]
nn = 14;
f[n_] := DivisorSum[n, MoebiusMu[#] # &]
n = nn;
Table[f[GCD[n, n - k + 1]], {k, 1, n}]
S = Sum[Sum[f[GCD[n, k]]*x^(n*h + k), {k, 1, n}], {h, 0, Infinity}]
integral =
Sum[Sum[f[GCD[n, k]]*x^(1 + k + h n)/(1 + k + h n), {k, 1,
n}], {h, 0, Infinity}];
Expand[FullSimplify[%]]
Expand[(-f[GCD[n, n - 1 + 1]])*
x*(1 - Hypergeometric2F1[1/n, 1, (n + 1)/n, x^n]) +
Sum[f[GCD[n, n - k + 1]]/k*x^k*
Hypergeometric2F1[k/n, 1, (n + k)/n, x^n], {k, 2, n}]]
%% - %
The general formula for the integral for any $n$ appears to be given by this Mathematica 14 program:
Clear[f, n, k, x, h, A, B]
nn = 30;
f[n_] := DivisorSum[n, MoebiusMu[#] # &]
n = nn;
Table[f[GCD[n, n - k + 1]], {k, 1, n}]
S = Sum[Sum[f[GCD[n, k]]*x^(n*h + k), {k, 1, n}], {h, 0, Infinity}]
integral =
Expand[Sum[
Sum[f[GCD[n, k]]*x^(1 + k + h n)/(1 + k + h n), {k, 1, n}], {h,
0, Infinity}]]
-f[GCD[n, n]] x +
f[GCD[n, n]]*x Hypergeometric2F1[1/n, 1, (n + 1)/n, x^n] +
Sum[f[GCD[n, k - 1]]/k*x^k*
Hypergeometric2F1[k/n, 1, (n + k)/n, x^n], {k, 2, n}]
%% - %
which returns 0 which means they should be equivalent. I found the general formula for the integral in a much simpler way by writing StringJoin["q", ToString[k]] and inspecting the result.
Apparently one gets long denominators for any sequence f that sums to zero:
Clear[f, n, k, x, h, A, B]
f = {3, -2, -7, 2, 3, -2, 1, -2, 2, 2}
Total[f]
n = Length[f];
S = Sum[Sum[f[[k]]*x^(n*h + k), {k, 1, n}], {h, 0, Infinity}]
integral =
Expand[Sum[
Sum[f[[k]]*x^(1 + k + h n)/(1 + k + h n), {k, 1, n}], {h, 0,
Infinity}]]
-f[[n]] x + f[[n]]*x Hypergeometric2F1[1/n, 1, (n + 1)/n, x^n] +
Sum[f[[k - 1]]/k*x^k*Hypergeometric2F1[k/n, 1, (n + k)/n, x^n], {k,
2, n}]
%% - %
f need not sum to zero but those sequences give uninteresting denominators.
