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I'm a bit stumped on a problem. The problem is as follows:

Suppose $f(z)$ is entire and $|f(x+iy)|\leq ae^{-bx^2+cy^2}$ for $a,b,c>0$. If $\hat{f}(z)=\int_{-\infty}^\infty f(x)e^{-2\pi ixz}dx$, then $|\hat{f}(\zeta+i\eta)|\leq a'e^{-b'\zeta^2+c'\eta^2}$.

There's a hint that tells me to assume $\zeta>0$ and change the contour of integration to $x-iy$ for $y>0$ fixed, and $x$ between $-\infty$ and $\infty$ to show that $\hat{f}=O(e^{-b'\zeta^2})$. Then it follows that $\hat{f}(\zeta)=O(e^{-2\pi y\zeta}e^{cy^2})$, and the result follows if we choose $y=d\zeta$ with $d$ a small constant.

I've tried using residues to evaluate $\hat{f}$, but with no luck. I also don't know how to use the hint-- if there's anybody who could help elucidate or tell me what to do here, that would be greatly appreciated, as I'm quite confused. Thanks in advance :)

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Your idea with the residue theorem was fine you should just note that the function does not have any poles (because it is entire). Thus you can deform the contour of integration as you please (keeping the end-points fixed) and you won't change the value of the integral.

In particular following your hint you should move the integration contour away from the real axis along $x-i y$ with $y>0$ fixed, i.e., $$\hat{f}(z)=\int_{-\infty}^\infty f(x)e^{-2\pi ixz}dx = \int_{-\infty}^\infty f(x-i y)e^{-2\pi i(x-i y)z}dx = \cdots$$

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Got it! Thanks! –  Jay C Mar 22 '12 at 6:03

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