Analytically continue a function with Euler product I would like to estimate the main term of the integral
$$\frac{1}{2\pi i} \int_{(c)} L(s) \frac{x^s}{s} ds$$
where $c > 0$, $\displaystyle L(s) = \prod_p \left(1 + \frac{2}{p(p^s-1)}\right)$.

Question: How to estimate the integral? In other words, is there any way to analytic continue this function?

The function as stated converges for $\Re s > 0$, but I'm not sure how to extend it past $y$-axis. Thanks!
 A: Let $\rho(d)$ count the number of solutions $x$ in $\frac{Z}{dZ}$,  to $x^2\equiv \text{-1 mod d}$, then we have
$$\sum_{n\leq x}d(n^2+1)=2x\sum_{n\leq x}\frac{\rho(n)}{n}+O(\sum_{n\leq x}\rho(n))$$
By multiplicative properties of $\rho(n)$ we have,
$$\rho(n)=\chi(n)*|\mu(n)|$$
Where $\chi(n)$ is the non principal character modulo $4$
Which allows us to estimate,
$$\sum_{n\leq x}\frac{\rho(n)}{n}=\sum_{n\leq x}\frac{\chi(n)*|\mu(n)|}{n}=\sum_{n\leq x}\frac{|\mu(n)|}{n}\sum_{k\leq \frac{x}{n}}\frac{\chi(k)}{k}=\sum_{n\leq x}\frac{|\mu(n)|}{n}(L(1,\chi)+O(\frac{n}{x}))=L(1,\chi)\sum_{n\leq x}\frac{|\mu(n)|}{n}+O(1)=\frac{\pi}{4}\sum_{n\leq x}\frac{|\mu(n)|}{n}+O(1)=\frac{\pi}{4}(\frac{6}{\pi^2}\ln(x)+O(1))$$
So that, $$\sum_{n\leq x}\frac{\rho(n)}{n}=\frac{3}{2\pi}\ln(x)+O(1)$$
Also note that, $$\sum_{n\leq x}\rho(n)=\sum_{n\leq x}{\chi(n)*|\mu(n)|}=\sum_{n\leq x}|\mu(n)|\sum_{k\leq \frac{x}{n}}\chi(k)\leq\sum_{n\leq x}|\mu(n)|\leq x$$
So we have, $$\sum_{n\leq x}\rho(n)=O(x)$$
Which gives,
$$\sum_{n\leq x}d(n^2+1)=\frac{3}{\pi}x\ln(x)+O(x)$$
