I took the theta function
$$\theta(\tau)= \sum_{n\in\Bbb Z} e^{\pi in^2\tau}$$
and made the substitution $\tau=\frac{1}{\log t}.$ This gave me
$$ \psi(t)=\sum_{n\in \Bbb Z} e^{\frac{\pi i n^2}{\log t}}. $$
Then I took the Mellin transform of $\psi(t)$ which I wrote down as
$$ \hat \Psi(r)=\int_{I=(0,1)} \psi(t)t^{r-1}dt $$
This integration is non-trivial so I used wolfram alpha to obtain
$$ \hat \Psi(r)= \sum_{n\in \Bbb Z} 2\sqrt{\frac{\pi i n^2}{r}}K_1\big( 2\sqrt{\pi i n^2 r} \big)$$
where $K_1$ is the modified Bessel function of the second kind.
Since the Mellin transform of $\theta(\tau)$ function yields the completed Riemann zeta function, I concluded that $\hat \Psi(r)$ is the "completed Bessel function."
However I don't know how to make this equation true
$$ \hat \Psi(r):= \Lambda(r) \Psi(r) $$
Analogous to the well known equation
$$ \hat \zeta(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s) $$
such that
$$ \hat \Psi(r)=\hat \Psi(1-r). $$
In my attempt to resolve this I skipped Poisson summation knowing that $\theta$ satisfies a functional equation, and I used $\tau=\frac{1}{\log t}$ to obtain
$$ \psi(t)=\sqrt{\log t}~\psi\big(\log t\big).$$
Therefore under the Mellin transform I took of $\psi(t)$ this implies the functional equation
$$ \hat \Psi(r)=\hat \Psi(1-r). $$
But I am still stuck on obtaining the factor $\Lambda(r)$ which makes this true.
What is the correct factor $\Lambda(r)$?
I think $$\Lambda(r)=2\sqrt{\frac{\pi}{r}}K_1(2\sqrt{\pi r}) $$
But it is just a guess.