The Riemann $\Xi(z)$ function is defined as $$\Xi(z)=2\int_1^\infty A(x)x^{-1/4}\cos\left((1/2)z\ln x\right)dx$$
where $A(x)$ is given in terms of Jacobi theta function $$\psi(x)=(1/2)(\theta_3(0,e^{-\pi x})-1)=\sum_{n=1}^{\infty}\exp(-n^2\pi x)$$ by $$A(x)=2\frac{d}{dx}\left(x^{3/2}\psi'(x)\right)=\left(3\psi'(x)+2x\psi''(x)\right)x^{1/2}\qquad\tag{0}$$ $$A(x)=\sum_{n=1}^\infty (2n^4\pi^2x-3n^2\pi)x^{1/2}\exp(-n^2\pi x)$$
The $\psi(x)$ satisfies the relation (due to Jacobi) $$2\psi(x)+1=(1/x)^{1/2}(2\psi(1/x)+1)\qquad\tag{1}$$
Because $\theta_3(0,e^{-\pi x})$ is a modular form of weight $1/2$: $$\theta_3(0,e^{-\pi x})=(1/x)^{1/2}\theta_3(0,e^{-\pi/x})$$
Din (arXiv:1009.2989) mentioned that A(x) satisfies relation
$$A(x)=(1/x)^{3/2}A(1/x)\qquad\tag{2}$$
Thus $A(x)$ is a modular form of weight of $3/2$
I am looking for a derivation or a reference of derivation from (1) to (2).
Thanks- mike
EDIT: Dimitrov and Rusev (EAST JOURNAL ON APPROXIMATIONS,Volume 17, Number 1 (2011), 1-108 ZEROS OF ENTIRE FOURIER TRANSFORMS) mentioned that (1) implies the following relation: $$\left(3\psi'(x)+2x\psi''(x)\right)x^{5/4}=\left(3\psi'(1/x)+(2/x)\psi''(1/x)\right)(1/x)^{5/4}\qquad\tag{3}$$ Comparing (3) and (0), we obtain (2).
I have not figured out why (1) implies (3).
A friend of mine informed the tricks to prove that (1) implies (3).
Here are the steps.
From (1) we have: $$\psi(1/x)=(1/2)((2\psi(x)+1)x^{1/2}-1)\qquad\tag{4}$$
Thus $$\psi'(1/x)=\frac{d \psi((1/x))}{d(1/x)}=\left(\frac{d(1/x)}{dx}\right)^{-1}\frac{d \psi((1/x))}{dx}=\left(-\frac{1}{x^2}\right)\frac{d}{dx}\psi((1/x))\qquad \tag{5}$$ and
$$\psi''(1/x)=\left(\frac{1}{x^2}\right)\frac{d}{dx}\left[\left(\frac{1}{x^2}\right)\frac{d}{dx}\psi((1/x))\right]\qquad \tag{6}$$
Substituting (4) into the RHS of (5) and (6) and then substituting the results into the RHS of (3), we found out that it equals to LHS of (3).