As from the title, I am currently dealing with this sum
$\sum_{n=0}^\infty \frac{1}{(a+n^2)^p}$
in particular with $p=1/2,3/2,...$ (but once solved for $p=1/2$ one can derive wrt $a$ and find the others), and real positive $a$. I would like to understand which kind of zeta function is it (Shintani's zeta??) and how to regularise it (Mathematica tells me it does not converge). Any good reference dealing with this kind of sums?
As I said in the comment, $$F_a(s) = \sum_{n=0}^\infty (n^2+a)^{-s}, \qquad Re(s) > 1/2$$ Has an analytic continuation in term of the Riemann zeta function :
$$F_a(s) = \sum_{n= 0}^{A-1} (n^2+a)^{-s}+\sum_{k=0}^\infty {-s \choose k} a^k \underset {\underset{\underset{\displaystyle s \in \mathbb{C}\setminus \{1/2-k,k\in \mathbb{N}\}}{}}{ }}{\left(\zeta(2k+2s)-\sum_{n=1}^{A-1} n^{-2k-2s}\right),}\qquad (1)$$
where $A > |a|^{1/2}$
for $n > |a|^{1/2}$ : $(n^2+a)^{-s} = n^{-2s}(1+\frac{a}{n^2})^{-s} = n^{-2s}\sum_{k=0}^\infty {-s \choose k} a^k n^{-2k}$. thus, with $A= \lfloor \, |a|^{1/2} \, \rfloor+1$, on $Re(s) > 1/2$ where everything converges absolutely :
$$\begin{eqnarray}F_a(s) -\sum_{n= 0}^{A-1} (n^2+a)^{-s} &=& \sum_{n= A}^\infty (n^2+a)^{-s} \\ &=& \sum_{n= A}^\infty \sum_{k=0}^\infty {-s \choose k} a^k n^{-2k-2s} \\ &=& \sum_{k=0}^\infty {-s \choose k} a^k \sum_{n= A}^\infty n^{-2k-2s} \\ &=& \sum_{k=0}^\infty {-s \choose k} a^k \left(\zeta(2k+2s)-\sum_{n=1}^{A-1} n^{-2k-2s}\right) \end{eqnarray}$$
now since the radius of convergence of $(1+x)^{-s}= \sum_{k=0}^\infty {-s \choose k} x^k$ is $\ge 1$ for every $s$, it means ${-s \choose k} = \mathcal{O}((1+\epsilon)^k)$, and since $\zeta(2k+2s)-\sum_{n=1}^{A-1} n^{-2k-2s} = \mathcal{O}(A^{-2k})$, we have that $(1)$ converges compactly for every $s \in \mathbb{C}$ where the summand are analytic,
and hence it defines the analytic continuation of $F_a(s)$ that is meromorphic with poles at $s = 1/2-k$
The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$.
For $a\gg 1$ you can then approximate the sum by an integral, to arrive at
$$I_{a,p}=\int_0^\infty \frac{1}{(a+x^2)^p}dx=\frac{\sqrt{\pi a } \Gamma (p-1/2)}{2 a^p\Gamma (p)},\;\;\text{for}\;\;p>1/2.$$
This compares quite well with the sum
$$S_{a,p}=\sum_{n=0}^\infty\frac{1}{(a+n^2)^p}$$
as you can see from the plot where both $I_{a,p}$ (blue) and $S_{a,p}$ (orange) are plotted as a function of $a$ for $p=0.51$.
This is for $p=1/2+\epsilon$. For the larger values of $p$ you mention no regularization is needed.
