A posted exercise is from Complex Analysis, Stein
This exercise generalizes some of the properties of $e^{-\pi{x^2}}$ related to the fact that it is its own Fourier transform.
Suppose $f(z)$ is an entire function that satisfies $$|f(x+iy)|\le ce^{-a{x^2}+b{y^2}}$$ for some $a,b,c\gt0.$
Let $$\hat f(\zeta)=\int^{\infty}_{-\infty}f(x)e^{-2\pi{i}x\zeta}dx.$$
Then, $\hat f$ is an entire fuction of $\zeta$ that satisfies $$|\hat f(\xi+i\eta)|\le c^{'}e^{-{a^{'}\xi^2+b^{'}\eta^2}}$$ for some $a^{'},b^{'},c^{'}\gt0.$
I tried to show $\hat f(\xi)=O(e^{-a^{'}\xi^2})$ that I take $y=\xi$ and consider the horizontal strip. Clearly, $f(z)$ is entire and satisfying the decay condition hence $\hat f(\xi)=O(e^{-2\pi{y}\xi}e^{b{y^2}})$ by using the contour integration.
But I cannot make it sure explicitly how to use the contour integration $-R\lt{x}\lt{R},y=\xi$ as $R$ goes to $\infty$.
Deform the contour so that
$$\hat f(\zeta)=\int_{-\infty}^\infty f(x+iy)e^{-i2\pi(x+iy)(\xi+i\eta)}\,dx$$
Then note that
$$\begin{align} |\hat f(\zeta)|&\le\int_{-\infty}^\infty |f(x+iy)|e^{2\pi(x\eta+y\xi)}\,dx\\\\ &\le ce^{by^2+2\pi y \xi}\int_{-\infty}^\infty e^{-ax^2+2\pi x\eta}\,dx\\\\ &\le ce^{by^2+2\pi y \xi}\,e^{\frac{\pi^2}{a}\eta^2} \int_{-\infty}^\infty e^{-a\left(x-\frac{\pi \eta}{a}\right)^2}\,dx\\\\ &=\sqrt{\frac{\pi}{a}}\,c \,e^{\frac{\pi^2}{a}\eta^2}\,e^{by^2+2\pi y \xi} \end{align}$$
First, choose $y=-\alpha \xi$ where we also choose $\alpha$ so that it satisfies the inequality $-2\pi/b< \alpha <0$.
Then, denoting
$$\begin{align} a'&=-2\pi \alpha -b\alpha^2\\\\ b'&=\frac{\pi^2}{a}\\\\ c'&=\sqrt{\frac{\pi}{a}}\,c \end{align}$$
we have
$$|\hat f(\zeta)|\le c'e^{-a'\xi^2+b'\eta^2}$$
as was to be shown!