Generalized constraint identities for odd divisor functions

In fact, the constraint follows directly from the functional equation of the completed Riemann zeta function; the restriction that $a$ be odd is unnecessary.

Put

$$g(s)=\pi^{-s/2}(s-1)\Gamma\!\left(\frac{s}{2}+1\right), \qquad \xi(s)=g(s)\zeta(s).$$

Then

$$\xi(s)=\frac12s(s-1)\pi^{-s/2} \Gamma\!\left(\frac s2\right)\zeta(s)$$

is entire and satisfies

$$\xi(1-s)=\xi(s).$$

Initially choose $\sigma>1+\max\{0,\operatorname{Re}a\}$. The absolutely convergent Dirichlet series

$$\sum_{n\ge1}\frac{\sigma_a(n)}{n^s} =\zeta(s)\zeta(s-a), \qquad \sum_{n\ge1}\frac{\sigma_a(n)}{n^{s+a}} =\zeta(s+a)\zeta(s)$$

and the exponential decay of the gamma factors justify interchanging summation and integration. The proposed left-hand side therefore becomes

$$-\int_{\operatorname{Re}s=\sigma} \xi(s)\xi(s-a)(-s)^k\,ds + \int_{\operatorname{Re}s=\sigma} \xi(s)\xi(s+a)(s-1)^k\,ds.$$

Define the entire function

$$F(s)=\xi(s)\xi(s+a)(s-1)^k.$$

By the functional equation,

$$F(1-s)=\xi(s)\xi(s-a)(-s)^k.$$

The substitution $w=1-s$, with the induced reversal of the vertical path canceled by $ds=-dw$, gives

$$\int_{\operatorname{Re}s=\sigma}F(1-s)\,ds = \int_{\operatorname{Re}w=1-\sigma}F(w)\,dw.$$

Since $F$ is entire and decays exponentially along horizontal edges in every fixed vertical strip, Cauchy's theorem yields

$$\int_{\operatorname{Re}w=1-\sigma}F(w)\,dw = \int_{\operatorname{Re}s=\sigma}F(s)\,ds.$$

The two terms consequently cancel, proving the asserted identity for every integer $k\ge0$.

For any other contour lying to the right of the poles in the original statement, shift each individual summand's gamma-product contour to a line satisfying $\sigma>1+\max\{0,\operatorname{Re}a\}$. No pole is crossed, and the horizontal integrals vanish by Stirling's estimate. The individual integrals are therefore unchanged, so the same identity holds on every admissible contour. In particular, it holds for every positive odd integer $a$ as conjectured.