Question
Can this relationship be simplified further? $ \def\lf{\left\lfloor} \def\rf{\right\rfloor} $
$$ \sum_{i=1}^n \mu(i)(\lf \frac{n}{i}\rf \ln(\lf \frac{n}{i}\rf + 1) + \frac{1}{2}\ln(\lf \frac{n}{i}\rf + 1) + \lf \frac{n}{i} \rf \ln(i) + \int_0^\infty \frac{2 \tan^{-1}(t/(\lf \frac{n}{i}\rf + 1))}{e^{2 \pi t} -1}dt)=1 +(1-\frac{1}{2}\ln(2 \pi)) M(n) $$
Derivation
Using the Mobius inversion formula on the factorial function and the floor function I obtained the following identity: $$ \prod_{i=1}^n (i^{\lf \frac{n}{i} \rf} \lf \frac{n}{i}\rf !)^{\mu(i)} = 1$$
$$ \implies \sum_{i=1}^n \mu(i)(\ln\lf \frac{n}{i}\rf ! + \lf \frac{n}{i} \rf \ln(i)) =0 $$
$$ \implies \sum_{i=1}^n \mu(i)(\ln\Gamma(\lf \frac{n}{i}\rf + 1) + \lf \frac{n}{i} \rf \ln(i)) =0 $$
Using Stirling's formula
$$ \implies \sum_{i=1}^n \mu(i)(\lf \frac{n}{i}\rf + 1)(\ln(\lf \frac{n}{i}\rf + 1) -(\lf \frac{n}{i}\rf + 1) +\frac{1}{2}\ln(2 \pi)- \frac{1}{2}\ln((\lf \frac{n}{i}\rf + 1) + \lf \frac{n}{i} \rf \ln(i)) + \int_0^\infty \frac{2 \tan^{-1}(t/(\lf \frac{n}{i}\rf + 1))}{e^{2 \pi t} -1})=0 $$
Expanding terms and using the mobius inversion formula on $ \sum 1 = n \implies \sum \mu(i) \lf \frac{n}{i}\rf = 1 $
Where $M(n)$ is the Mertens function:
$$ \implies \sum_{i=1}^n \mu(i)(\lf \frac{n}{i}\rf \ln(\lf \frac{n}{i}\rf + 1) + \frac{1}{2}\ln(\lf \frac{n}{i}\rf + 1) + \lf \frac{n}{i} \rf \ln(i) + \int_0^\infty \frac{2 \tan^{-1}(t/(\lf \frac{n}{i}\rf + 1))}{e^{2 \pi t} -1}dt)=1 +(1-\frac{1}{2}\ln(2 \pi)) M(n) $$
P.S: I'm only a physics undergraduate