Theorem 1: H-S transform in real domain (scalar)
$e^{x^2}=\sqrt{\frac{\eta}{2\pi}}\int \exp \left({-\frac{\eta}{2}z^2+\sqrt{2\eta}xz}\right)\text{d}z, \quad \forall x,\eta$
$proof$: Given the Gaussian distribution $\mathcal{N}(x|\sqrt{\frac{2}{\eta}}z,\frac{1}{\eta})$, we have following formula , thanks to the normalization of PDF,
$\int \sqrt{\frac{\eta}{2\pi}}\exp \left[-\frac{\eta}{2}(x-\sqrt{\frac{2}{\eta}}z)^2\right]\text{d}x=1$
Expand the formula above, we get
$\int \sqrt{\frac{\eta}{2\pi}}\exp \left(-\frac{\eta}{2}x^2+\sqrt{2\eta}xz\right)\text{d}x=e^{z^2}$
Interchanging the $x$ and $z$, we have finished the proof.

Theorem 2: H-S transform in real domain (scalar)
$e^{-x^2}=\sqrt{\frac{\eta}{2\pi}}\int \exp \left({-\frac{\eta}{2}z^2+j\sqrt{2\eta}xz}\right)\text{d}z, \quad \forall x,\eta$
where $j=\sqrt{-1}$.

$proof$: Similarly as the proof of theorem 1, we also use the normalization of Gaussian PDF. Given the Gaussian distribution as below
$\mathcal{N}\left({x|j\sqrt{\frac{2}{\eta}}z,\frac{1}{\eta}}\right)=\sqrt{\frac{\eta}{2\pi}}\exp \left[{-\frac{\eta}{2}(x-j\sqrt{\frac{2}{\eta}}z)^2}\right]$
Using the normalization of PDF, we have
$\int \sqrt{\frac{\eta}{2\pi}}\exp \left[{-\frac{\eta}{2}x^2+j\sqrt{2\eta}xz}\right]\text{d}x=e^{-z^2}$
Interchanging $x$ and $z$ yields the theorem 2.

Theorem 3:
$e^{\|\boldsymbol{x}\|^2}=\int \left({\frac{\eta}{2\pi}}\right)^{N/2}\exp \left({-\frac{\eta}{2}\boldsymbol{z}^T\boldsymbol{z}+\sqrt{2\eta}\boldsymbol{x}^T\boldsymbol{z}}\right)\text{d}\boldsymbol{z}$
$proof$: Given the vector Gaussian distribution
$\mathcal{N}\left({\boldsymbol{x}|\sqrt{\frac{2}{\eta}}z,\frac{1}{\eta}\mathbf{I}}\right)=\left({\frac{\eta}{2\pi}}\right)^{N/2}\exp \left[{-\frac{\eta}{2}\|\boldsymbol{x}-\sqrt{\frac{2}{\eta}}\boldsymbol{z}\|^2}\right]$
Using the normalization of Gaussian distribution, we have
$\int \left({\frac{\eta}{2\pi}}\right)^{N/2} \exp \left[{-\frac{\eta}{2}\boldsymbol{x}^T\boldsymbol{x}+\sqrt{2\eta}\boldsymbol{x}^T\boldsymbol{z}}\right]\text{d}\boldsymbol{x}=e^{\boldsymbol{z}^T\boldsymbol{z}}$
Interchanging the $\boldsymbol{x}$ and $\boldsymbol{z}$ yields theorem 3.

Theorem 4:
$e^{-\|\boldsymbol{x}\|^2}=\int \left({\frac{\eta}{2\pi}}\right)^{N/2}\exp \left({-\frac{\eta}{2}\boldsymbol{z}^T\boldsymbol{z}+j\sqrt{2\eta}\boldsymbol{x}^T\boldsymbol{z}}\right)\text{d}\boldsymbol{z}$
$proof$: see the proof of theorem 2 and theorem 3.