<s>[INST] Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x  \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
 [/INST] As Paul Enta already answered, the integral can be exactly computed.
$$J=\int_z^\infty \exp\left({- \frac { \pi^2}{16}(y-x)^2}\right) \,d y=\frac{2 }{\sqrt{\pi }}\left(1+\text{erf}\left(\frac{\pi}{4}   (x-z)\right)\right)$$
$$I=\int \frac{\cos \left(\frac{\pi  x}{2 z}\right)}{z}J \,dx=\frac{2 e^{-\frac{1}{z^2}}}{\pi ^{3/2}}\left(\text{erf}\left(\frac{\pi  z (z-x)+4 i}{4 z}\right)-\text{erf}\left(\frac{\pi  z (x-z)+4 i}{4 z}\right) \right)+\frac{4 \sin
   \left(\frac{\pi  x}{2 z}\right)}{\pi ^{3/2}} \left(1+\text{erf}\left(\frac{\pi  (x-z)}{4} \right)\right)$$ Now, using the bounds, the result of the given definite integral is
$$\frac{2 \left(4-2 \text{erf}\left(\frac{\pi  z}{2}\right)-i e^{-\frac{1}{z^2}}
   \left(\text{erfi}\left(\frac{1}{z}-\frac{i \pi 
   z}{2}\right)-\text{erfi}\left(\frac{1}{z}+\frac{i \pi 
   z}{2}\right)\right)\right)}{\pi ^{3/2}}$$ and expanding for large values of $z$, this gives
$$\sim \frac{4}{\pi ^{3/2} z^2}\left(1-\frac{1}{2 z^2}+\frac{1}{6 z^4}+O\left(\frac{1}{z^6}\right) \right)+\frac{32 e^{-\frac{1}{4} \pi ^2 z^2}}{\pi ^5 z^5}\left(1-\frac{12}{\pi ^2 z^2}+\frac{4 \left(45-\pi ^2\right)}{\pi ^4
   z^4}+O\left(\frac{1}{z^6}\right) \right)$$
</s>