I'm trying to show the following identity:
If $f=\sum_{n\geq 1}a(n)q^n \in S_k$ is a normalized Heckeeigenform, where $k$ can be written as a sum of two even numbers $k= k_1+k_2$, and $L(f,s)=\sum_{n\geq 1}a(n)n^{-s}$ its corresponding $L$-function, is it then true that:
$\sum_{n\geq 1}\dfrac{a(n)\sigma_{k_1-1}(n)}{n^s} = \dfrac{L(f,s)L(f,s-k_1+1)}{\zeta(2s-k_1-k+2)}$ ?
I am actually only interested in showing $\sum_{n\geq 1}\dfrac{a(n)\sigma_{k_1-1}(n)}{n^{k-1}} = \dfrac{L(f,k-1)L(f,k_2)}{\zeta(k_2)}$, but the plan was to compare the factors in the Euler product expansions of the first equality. Unfortunately, I seem to be missing the point, although this is said to be an easy exercise in the material I am currently reading.
Any help would be very appreciated! Also any references to papers or books dealing with this kind of L-functions would be great!
Thanks!
The key point is that both $a(n)$ and $\sigma_{k_1 - 1}(n)$ are multiplicative functions, and so \[\sum_{n = 1}^{\infty} \frac{a(n) \sigma_{k_1 - 1}(n)}{n^s} = \prod_p \sum_{m = 0}^{\infty} \frac{a(p^m) \sigma_{k_1 - 1}(p^m)}{p^{ms}}.\] On the other hand, \[\frac{L(s,f) L(s - k_1 + 1,f)}{\zeta(2s - k_1 - k + 2)} = \prod_p \frac{1 - p^{-(2s - k_1 - k + 2)}}{(1 - a(p) p^{-s} + p^{k - 1 - 2s}) (1 - a(p) p^{-(s - k_1 + 1)} + p^{k - 1 - 2(s - k_1 + 1)})}.\] So we must show that for every prime $p$, \[\sum_{m = 0}^{\infty} \frac{a(p^m) \sigma_{k_1 - 1}(p^m)}{p^{ms}} = \frac{1 - p^{-(2s - k_1 - k + 2)}}{(1 - a(p) p^{-s} + p^{k - 1 - 2s}) (1 - a(p) p^{-(s - k_1 + 1)} + p^{k - 1 - 2(s - k_1 + 1)})}.\] This is tedious but doable; the trick is to clear denominators of both sides and use the fact that \[a(p^{m + 1}) = a(p) a(p^m) - p^{k - 1} a(p^{m - 1})\] and that \[\sigma_{k_1 - 1}(p^m) = 1 + p^{k_1 - 1} + p^{2(k_1 - 1)} + \cdots + p^{m(k_1 - 1)} = \frac{p^{(m + 1)(k_1 - 1)} - 1}{p^{k_1 - 1} - 1}.\]