I was reading a paper on the moments of the Riemann zeta function where I have found the following equation.
Let $\sigma_{\alpha,\beta}(n)=\sum_{n_1n_2=n}n_1^{-\alpha}n_2^{-\beta}=n^{-\alpha-\beta}\sum_{n_1n_2=n}n_1^{\alpha}n_2^{\beta}$, where $\sigma_{\lambda}(n)=\sum_{d|n} d^{\lambda}$ and $\alpha,\beta$ are small complex numbers. Further we have
\begin{align*}
A_{\alpha,\beta,\gamma,\delta}(s) =\frac{\zeta(1+s+\alpha+\gamma)\zeta(1+s+\alpha+\delta)\zeta(1+s+\beta+\gamma)\zeta(1+s+\beta+\delta)}{\zeta(2+2s+\alpha+\beta+\gamma+\delta)},
\end{align*}
where $\alpha,\beta,\gamma,\delta$ are small complex numbers. Claim: $$\sum_{n=1}^{\infty}\frac{\sigma_{\alpha,\beta}(n)\sigma_{\gamma,\delta}(n)}{n^{1+s}}=A_{\alpha,\beta,\gamma,\delta}(s).$$
I can not understand how does the above equality hold.
Best Answer
Not a complete answer.
Using the product formula for $\zeta$ we get the product term for $p$ in the definition of $A$ is:
$$\frac{1-p^{-(2+2s+\alpha+\beta+\gamma+\delta)}}{\left(1-p^{-(1+s+\alpha+\gamma)}\right) \left(1-p^{-(1+s+\alpha+\delta)}\right) \left(1-p^{-(1+s+\beta+\gamma)}\right) \left(1-p^{-(1+s+\beta+\delta)}\right)}\tag1 $$
Also, $\sigma_{\alpha,\beta}$ and $\sigma_{\gamma,\delta}$ are both multiplicative, so the left side of your claim is the product of terms:
$$f_p(s)=\sum_{k=0}^{\infty} \frac{\sigma_{\alpha,\beta}(p^k)\sigma_{\gamma,\delta}(p^k)}{p^{k(1+s)}}\tag 2$$
You need to show (1) an (2) are equal.
Presumably, you need the calculation that:
$$\begin{align}\sigma_{\alpha,\beta}(p^k)&=\sum_{i=0}^{k}(p^i)^{\alpha}(p^{k-i})^\beta\\ &=\frac{p^{(k+1)\alpha}-p^{(k+1)\beta}}{p^\alpha-p^{\beta}} \end{align}$$
After that, things get even messier, but you should be able to get a formula for $f_p.$
If they are the same, the formula for (2) is going to look like a partial fraction form for the formula (1).