how to prove an approximate to Riemann Xi function having only real zeros

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I am searching for approximates to Riemann Xi function.

Riemann $\Xi(z)$ function is related to Riemann $\zeta(s)$ function via $$\Xi(z)=\xi(1/2+iz)=\xi(s)=(1/2)s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$$

Riemann Hypothesis is that all the roots of $\Xi(z)$ are real.

Let $$f(z)=\pi^{-1/4-iz/2}\left(2\gamma\left(5/4+iz/2,\pi/2\right)-2\gamma\left(5/4+iz/2,2\pi\right)-3\gamma\left(5/4+iz/2,\pi/2\right)+3\gamma\left(5/4+iz/2,2\pi\right)\right)\tag{1}$$ $$F(z)=f(z)+f(-z)\tag{2}$$

In (2) $\gamma(s,x)$ is the (lower) incomplete gamma function and $\gamma(s,\infty)=\Gamma(s)$. The lower incomplete gamma function admits series expansion like:

$$\gamma(s,x)=x^s e^{-x}\sum_{k=0}^{\infty}\frac{x^k}{s(s+1)\cdots (s+k)}=x^s e^{-x}\Gamma(s)\sum_{k=0}^{\infty}\frac{x^k}{\Gamma(s+k+1)}\tag{3}$$

Here is a contour plots of $Re(F(x+iy))=0$ (in blue color) and $Im(F(x+iy))=0$ (in orange color).Re(F(z))=0,Im(F(z))=0

This figure and other figures (not shown here) seemed to us that all the zeros of $F(z)$, at least in the strip $Im(z)<1/2$, are real.

I am looking for solutions to prove that all the zeros of $F(z)$, at least in the strip $Im(z)<1/2$, are real.

Any reference on methods to prove similar functions having only real zeros are welcomed as well.

Best regards- mike