Let $f$ be a Modular form/Maass form on $GL(2)$ with level $N$ and character $\eta$ and Fourier coefficients $a(n)$.


The Rankin-Selberg convolution
$$L(s,f\times\bar f)=\sum \frac{a(n)\overline{a(n)}}{n^s}$$ has a pole at $s=1$.




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My question: 


What's the analytic conductor/level for $L(s,f\times\overline{f})$? $N$ or $N^2$?


What's the functional equation?


$$L(s,f\times\overline f ) N^{s/2} \* \gamma =L(1-s,f\times\overline f ) N^{(1-s)/2} \* \gamma$$


or 


$$L(s,f\times\overline f ) N^{s} \* \gamma =L(1-s,f\times\overline f ) N^{(1-s)} \* \gamma$$



Functional equation and conductor for a Rankin-Selberg convolution
First, a few things about normalization. The expression $L(s,f \times \overline{f})$ needs to be multiplied by $\zeta(2s)$ in order to have a functional equation. Also, if $f$ is a holomorphic modular form of weight $k$, we should have the $n$th Fourier coefficient be $a(n) n^{\frac{k-1}{2}}$ in order for the functional equation to relate the value at $s$ and $1-s$.


In order to have a functional equation of the type you seek, it may be necessary to modify the Euler factors at the primes dividing $N$. Similarly, there is no completely uniform formula for the analytic conductor. (It often equals $N^{2}$, but it can be a proper divisor of $N^{2}$.) A high-brow reason for this is the following. There is a functorial lifting from $GL(2) \times GL(2) \to GL(4)$ that commutes with the local Langlands correspondence. If $\pi\_{p}$ is the local representation of $f$ at $p$, then the $L$-function of $\pi\_{p} \otimes \tilde{\pi}\_{p}$ need not be related to the naive local factor. For more detail about this, see the paper "$L$-series of Rankin type and their functional equations" by Winnie Li (published in Mathematische Annalen in 1979). Li deals with the holomorphic case.


The gamma factors depend on whether $f$ is holomorphic or not. If $f$ is holomorphic of weight $k$ the gamma factor is
$$ \pi^{-2s} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{s + k - 1}{2}\right) \Gamma\left(\frac{s + k}{2}\right)$$
while if $f$ is a Maass form with eigenvalue $\lambda = \frac{1}{4} + r^{2}$, the gamma factor is
$$ \pi^{-2s} \Gamma\left(\frac{s}{2}\right)^{2} \Gamma\left(\frac{s}{2} + ir\right) \Gamma\left(\frac{s}{2} - ir\right).$$


